LIBRARY 

OF  THE 

UNIVERSITY  OF  CALIFORNIA. 


Class 


ALTERNATING   CUEEENTS 

A  TEXT-BOOK  FOR  STUDENTS   OF   ENGINEERING 


BY 


C.  G.  LAMB,  M.A.,  B.Sc. 

CLARE   COLLEGE,   CAMBRIDGE 

ASSOCIATE  MEMBER   OF  THE   INSTITUTION   OF   ELECTRICAL   ENGINEERS 
ASSOCIATE  OF  THE   CITY  AND   GUILDS   OF   LONDON    INSTITUTE 


NEW   YORK: 

LONGMANS,   GREEN   &   CO 
LONDON:    EDWARD   ARNOLD 

[All  Rights  reserved] 


cf. 


PREFACE. 

number  of  text-books  dealing  with  the  present  subject  is 
already  so  large  that  a  few  words  are  necessary  to  explain 
the  reason  for  an  addition  to  that  number.  For  some  years  the 
author  has  been  engaged  in  lecturing  on  electrical  matters  to  the 
students  of  Engineering  at  the  Engineering  Laboratory,  Cambridge, 
and  he  has  experienced  a  difficulty  in  recommending  to  the  men 
a  suitable  book  on  this  subject.  In  the  course  of  study  for  the 
Mechanical  Sciences  Tripos  the  subject  of  Electrical  Engineering 
is  but  one  of  several,  and  consequently  a  student  has  not  the  time 
to  devote  to  a  proper  study  of  the  larger  books  already  available 
either  in  English  or  German,  and  the  smaller  books  scarcely  cover 
the  ground  with  which  the  course  deals.  Hence  it  was  felt  that 
a  compilation  of  the  more  important  points  was  desirable.  It  was 
also  hoped  that  such  a  compilation  might  possibly  be  of  some  use 
to  teachers  in  general.  In  such  a  book  as  the  present,  dealing 
merely  with  the  broad  outlines  of  the  subject,  little  that  is  not 
common  knowledge  can  be  embodied,  and  hence  very  few  refer- 
ences are  given.  To  the  student  such  references  are  merely 
distracting,  and  to  a  more  learned  person  (should  such  a  one 
honour  the  author  by  reading  the  book)  they  are  unnecessary. 

The  treatment  of  the  question  is  largely  based  on  the  use  of 
vectors,  supplemented  by  simple  analytical  methods  when  it  is 
desired  to  obtain  numerical  results.  The  symbolic  treatment 
has  been  found  by  the  author  to  appeal  to  a  very  limited  number 
of  students,  and  hence  has  not  been  used.  Throughout  the 
expressions  are  worked  out  in  general  terms — that  is,  no  attempt 
is  made  to  distinguish  in  the  formulae  whether  absolute  or 
practical  units  are  employed.  The  necessary  addition  to  all  the 
formulae  of  the  proper  factors  renders  the  expressions  unwieldy 
and  cumbrous :  the  student  who  has  proceeded  to  this  point  in  his 

203809 


IV  PREFACE 

subject  should  experience  no  difficulty  on  this  score,  and  numerical 
examples  are  worked  out,  which  will  serve  to  indicate  the  proper 
factors  that  should  be  used  in  each  case. 

It  may  strike  some  that  the  subjects  treated  of  differ  widely 
in  importance,  but  they  have  been  selected  chiefly  with  a  view  to 
the  elucidation  of  matters  of  principle  and  as  exemplifying  special 
points  of  theory. 

Throughout  the  book  there  is  no  descriptive  detail;  in  the 
author's  opinion,  such  detail  is  far  more  profitably  obtained  either 
by  actual  contact  with  drawing-office  work,  or  by  careful  perusal 
of  the  contemporary  technical  press :  further,  the  size  of  the  book 
necessarily  prohibited  the  inclusion  of  any  such  details. 

The  author  desires  to  thank  the  following  gentlemen :  Mr  G.  T. 
Bennett,  M.A.,  Emmanuel  College,  for  valuable  help  in  some  of 
the  geometrical  constructions,  Professor  B.  Hopkinson,  M.A., 
Trinity  College,  and  Mr  C.  E.  Inglis,  M.A.,  King's  College,  for 
communicating  certain  results,  Mr  T.  H.  Schoepf  for  figures  71 
and  72,  and  Mr  H.  Rottenburg,  M.A.,  King's  College,  for  kindly 
looking  over  the  proofs.  He  desires  also  to  thank  the  British 
Westinghouse  Company  for  lending  the  blocks  for  figures  51,  86, 
161,  162,  163,  and  174;  and  the  Cambridge  Scientific  Instrument 
Company  for  figures  93,  98,  99,  100,  101  and  103. 

In  a  book  of  this  description  errors  of  various  kinds  are  certain 
to  appear.  The  author  would  be  grateful  if  any  one  using  the 
book  would  communicate  such  errors  to  him. 

ENGINEERING  LABORATORY, 
CAMBRIDGE. 

August,  1906. 


CONTENTS. 

CHAP.  PAGE 

I.  SIMPLE  HARMONIC  PRESSURES  AND  CURRENTS      .        .        .          1 

II.  CURRENTS  DUE  TO  SIMPLE  HARMONIC  PRESSURE  ...        15 

III.  MEASUREMENT  OF  POWER 27 

IV.  THE  CHOKING  COIL  AND  TRANSFORMER         .        .        .        .        46 

V.  TRANSFORMER-EQUATIONS 66 

VI.  SPECIAL  FORMS  OF  TRANSFORMER .78 

VII.  LOSSES  IN  TRANSFORMERS 85 

VIII.      THE  SERIES  MOTOR 94 

IX.  THE  E.M.F.  OF  AN  ALTERNATOR.    HARMONICS        .        .        .102 

X.  EFFICIENCY  OF  ALTERNATORS.        .        .        .        .        .        .      135 

XL       POLYPHASE  CURRENTS 144 

XII.      MEASUREMENT  OF  POLYPHASE  POWER 152 

XIII       POLYPHASE  TRANSFORMATIONS 156 

XIV.  THE  ROTATING-FIELD  MOTOR 165 

XV.  THE  HEYLAND  CIRCLES 191 

XVI.  EQUATIONS  FOR  INDUCTION  MOTOR 203 

XVII.  MULTIPOLAR  MOTORS  ;  STARTING  ;  TESTING     .        .        .        .218 

XVIII.  THE  MONOPHASE  MOTOR        .        .        .        .        .        .        .224 

XIX.  METERS  OPERATING  BY  A  ROTATING  FIELD                                   230 


VI  CONTENTS 

CHAP.  PAGE 

XX.      ARMATURE  REACTION  BY  SYNCHRONOUS  IMPEDANCE      .        .  240 

XXI.  ARMATURE  KEACTION  IN  DETAIL 258 

XXII.  ALTERNATORS  IN  PARALLEL 273 

XXIII.  THE  SYNCHRONOUS  MOTOR 283 

XXIV.  THE  ROTARY  CONVERTER 310 

INDEX 323 


PREVIOUS   KNOWLEDGE   ASSUMED. 

General  elementary  theory  of  magnetism  including  the  idea  of 
self-induction,  the  magnetic  circuit,  hysteresis  and  the  laws 
of  eddy  currents. 

The  principles  involved  in  the  action  of  condensers. 
The  principles  of  direct  current  machines. 

The  elements  of  the  calculus,  including  the  definite  integrals  of 
simple  cases  of  the  circular  functions. 

The  elementary  properties  of  vectors. 


LIST    OF    SYMBOLS. 


ELECTROMOTIVE  FOKCE  OR  POTENTIAL  DIFFERENCE. 


Maximum 
Instantaneous 


E 
e,  € 


Virtual     g,  £ 
Direct  current 


V,  E 


Maximum 

Instantaneous 

Virtual 


CURRENTS. 


Power  component  <@p 
Wattless  component  ^q 
Direct  current  G 


POWER  OR  ENERGY. 


Power  in  general  W 
Lost  power  in 

general  Wi 

Mechanical  loss  Wf 

Ohmic  loss  W 

Hysteretic  loss  W 


Maximum  flux  <I> 

Instantaneous  flux  <f> 

Primary  flux  $x 

Secondary  flux  <£2 
Primary  leakage 
flux 


Eddy  current  loss  We 

Resistance  R,  r,  Rp, 

Reactance  S 

Impedance  / 
Angle  of  lag  or 

lead  X,  "a 


MAGNETIC  FLUXES. 


Secondary  leakage 
flux  <£s 

Magnetomotive 
force  A 

Reluctance  p 


Turns  of  wire  T 

Couple  P  Periodic  time 

Moment  of  inertia    /„  Time  in  seconds 

Revolutions  per 
second,  or  periods  n 

Angular  velocity  of  a  rotating  field 
Angular  velocity  of  a  rotating  body 
Slip  as  a  relative  angular  velocity- 
Slip  as  a  fractional  angular  velocity 


OF    THE 

UNIVERSITY 

OF 


CHAPTER   I. 

SOME   PROPERTIES   OF   SIMPLE   HARMONIC   QUANTITIES. 


Alternating  electromotive  force.  Let  a  coil  of  wire  be 
arranged  as  in  Fig.  1,  in  such  a  way  that  it  is  capable  of  rotation 
about  an  axis  and  has  its  ends  joined  to  two  rings  attached  to  the 
axis,  on  which  fixed  brushes  can  press.  Further  let  a  magnetic 
field  of  any  form  be  present,  the  direction  of  the  lines  of  force 
being  perpendicular  to  this  axis.  When  the  coil  has  a  position 
such  that  its  normal  is  in  the  direction  of  the  lines  of  force,  the 
total  flux  passing  through  it  will,  in  general,  be  a  maximum  and 
in  any  other  position,  will  be  a  function  of  the  angle  0,  between 
the  normal  to  the  coil  and  the  direction  of  the  lines  of  force ;  let 
this  function  be  denoted  by  f(B).  If  the  coil  be  rotated  the  flux 
will  undergo  a  rate  of  change  and  an  E.M.F.  will  be  generated  in 

the  coil  of  the  amount  given  by  the  expression  e  =  —  ~rf(0).    This 

7  7/1 

can  be  written  in  the  form  e  =  —  j/i/W  -~Ji-     If  tne  rotation  be 

made  with  the  uniform  angular  velocity,  o>,  we  can  replace  -j-  by  ay, 

and  hence  the  expression  becomes  in  this  case  e  =  —  a)f'(&).  If 
n  denote  the  number  of  rotations  per  second,  the  time  r  taken 

by  one  rotation  will   then  be   r  =  -.     This   time   is   called   the 

J  n 

periodic  time  of  the  rotation.  In  such  a  case  it  is  evident  that 
the  value  of  e  will  be  the  same  at  intervals  of  time  each  equal  to 
T  whatever  the  initial  position  of  the  coil  may  be. 

The  shape  of  the  curve  connecting  the  E.M.F.  and  the  time  will 
evidently  depend  on  the  form  of  the  curve  connecting  6  and  /(#). 
This  relation  depends  on  several  factors,  for  example  the  distri- 
bution in  space  of  the  flux  that  is  cut,  and  the  form  of  the  coil  both 
as  regards  the  shape  of  the  successive  turns  and  the  distribution 
of  them  in  space.  These  points  will  be  considered  in  Chapter  IX. 

L.  1 

Ml 


2  ALTERNATING   CURRENTS 

Simple  harmonic  E.M.F.  The  simplest  case  is  afforded  by 
a  uniform  distribution  of  flux  and  a  coil  with  all  the  turns 
practically  concentrated  in  one  place.  In  this  case  we  can  easily 
evaluate  the  expression  for  /(#).  Let  a  denote  the  area  of  one 
turn  of  the  coil,  T  the  number  of  turns,  and  let  the  field  in  which 
the  coil  rotates  be  such  as  to  produce  a  flux  of  B  lines  of  force 
per  square  centimetre.  If  we  reckon  the  position  angle  from  the 


Direction 
of  Motion 


Fig.  1. 

place  where  the  normal  to  the  coil  and  the  direction  of  the  flux 
coincide,  as  mentioned  before,  it  is  evident  that  the  relation 
between  the  flux,  </>,  that  passes  through  the  coil  and  its  position 
is  given  by  the  expression  </>  =  a.B  .  cos  6,  hence  in  this  case  the 
relation  between  6  and  the  E.M.F.,  e,  at  any  instant  will  be  given  by 
e  =  aBTa)  .  sin  0. 

For  a  definite  angular  velocity  we  can  put  E  for  the  maximum 
value  of  this  E.M.F.  and  we  then  get 

e  =  E  sin  0 


where  E  =  aBTa)  =  2irn  .  aBT. 

In  this  case  the  E.M.F.  is  said  to  be  a  simple  harmonic  one,  and 
can  then  be  represented  by  any  of  the  methods  usually  employed 
for  representing  a  simple  harmonic  quantity.  The  principal 
methods  are  as  follows  (see  Fig.  2). 

(1)  By  the  trace  of  sine  or  cosine  curve  whose   maximum 
ordinate  is  E. 

(2)  By  the  projection  of  a  vector  of  constant  length  E  on  any 
line,  either  the  vector  or  the  line  being  considered  to  be  rotating 
at  the  constant  angular  velocity  &>.     If  the  vector  be  considered 
as  rotating,  positive  direction   of  the   quantity  can   be  assumed 
to  correspond  to  the  vector  pointing  say,  upwards,  and  negative 
values  to  the  opposite  direction  of  pointing  :  when  the  line  rotates, 
the  sign  of  the  quantity  will  depend  on  which  side  of  this  line  the 
projection  falls. 


SIMPLE   HARMONIC   QUANTITIES  3 

(3)  By  a  Zeuner  Diagram  consisting  of  two  circles  of  the 
diameter  E  placed  with  the  diameters  in  a  line  and  touching 
at  the  common  point,  as  in  Fig.  2.  When  any  line  rotates 
about  the  centre  with  the  angular  velocity  co  the  part  intercepted 
by  the  circles  is  evidently  proportional  to  the  sine  of  the  angle 


Fig.  2. 

reckoned  from  the  common  tangent,  and  thus  gives  the  instan- 
taneous value  of  the  varying  quantity.  Each  of  these  methods 
has  its  special  advantages  for  special  cases;  the  first  two  are, 
however,  by  far  the  most  important  in  the  consideration  of 
electrical  matters. 

Since  the  rotation  of  the  coil  takes  place  with  constant  angular 
velocity,  it  is  a  matter  of  indifference  whether  the  independent 
variable  be  taken  as  the  angle  or  the  time.  The  one  can  always 
be  converted  into  the  other  if  it  be  remembered  that  the  time  r 
corresponds  to  a  complete  rotation  of  the  coil,  or  to  the  time  taken 
by  the  coil  to  return  to  its  original  position,  that  is,  to  turn 
through  the  angle  2?r.  Hence  if  0  be  the  angle  through  which 
the  coil  has  turned  in  the  time  I  we  have  the  relation 

T 

The  direction  in  which  the  time  is  reckoned  must  be  carefully 
borne  in  mind.  When  the  quantities  are  represented  by  sine 
curves  the  axis  of  x  can  be  taken  either  to  represent  the  time 
or  the  angle  as  we  have  seen :  the  positive  direction  of  time  will 
be  taken  to  correspond  with  x  increasing  in  value,  that  is  if  a^  and 
x*  are  two  values  of  the  abscissa,  the  latter  being  the  greater,  the 
events  corresponding  to  the  latter  value  will  be  considered  to  have 
occurred  subsequently  to  those  corresponding  to  the  first.  Again 
in  the  second  or  third  case  we  can  let  the  rotation  take  place 
either  clockwise  or  in  the  opposite  direction,  in  general  it  will  be 
taken  that  counter-clockwise  turning  corresponds  to  the  efflux 
of  time.  Thus  if  three  lines  be  drawn  as  in  Fig.  3,  the  rotation 

1—2 


ALTERNATING   CURRENTS 


taking  place  as  shown  by  the  arrow,  the  events  corresponding  to 
the  position  OB  will  be  considered  as  taking  place  after  those 


Fig.  3. 

corresponding  to  OA  while  in  the  same  way  the  events  cor- 
responding to  OC  precede  those  corresponding  to  OA  ;  this  is 
expressed  by  saying  that  OB  lags  after  OA  and  OC  leads  on  OA. 
The  case  where  the  varying  quantities  are  taken  as  being 
simple  harmonic  in  nature  is  the  one  usually  considered  most  fully 
as  the  relations  can  be  easily  treated  either  by  simple  analysis  or 
by  one  of  the  graphical  representations  mentioned  above.  We 
shall  see  later  on,  that  any  alternating  quantity  can  be  considered 
as  made  up  of  a  series  of  simple  harmonic  ones  the  successive 
periodic  times  of  which  diminish  in  the  ratio  of  the  natural 
numbers,  and  hence  the  most  complicated  case  can  be  treated 
as  a  sum  of  such  quantities.  It  is  thus  very  important  to  consider 
this  case  first. 

Mean  Value.  It  is  evident  that  if  e  =  E  sin  6  the  mean 
value  of  e  is  zero  over  the  complete  period,  and  in  this  case  it  is 
usual  to  take  as  the  mean  value  the  mean  for  one  half  of  the 
curve,  from  one  zero  value  of  the  ordinate  to  the  next.  In  the 
present  case  this  becomes 


mean«  =  -|   sin  0.d0  =  -  \  -  cos  6  \   =  -E  =  0'637E. 

7T 


~\n      9 

M    =-E  =  0' 

Jo         7T 


Virtual  Value.  A  much  more  important  quantity  is  that 
known  as  the  virtual  value  of  e.  In  all  cases  we  know  that  the 
rate  of  generation  of  heat  depends  on  the  square  of  a  current  or 
pressure,  and  hence  that  the  mean  rate  of  production  of  heat  will 
depeud  on  the  mean  value  of  this  square.  The  square  root  of 
this  mean  square  is  called  the  virtual  value  of  the  corresponding 
quantity,  and  we  will  denote  it  by  the  letter  S.  In  this  case,  then, 
we  have 

E2  f27r  E2  r2ir  E2  E2 

£'2  =  ^'-        sin20.d0= —      (1  -cos  20).d0=  — .  2-7T=  -— . 
27rJo  47rJo  4-7T  2 

Hence  g=  —  or  0'707E. 

Vz 


SIMPLE   HARMONIC   QUANTITIES  5 

This  relation  is  very  important ;  we  will  later  on  discuss  the  mean 
and  virtual  values  of  alternating  quantities  which  are  non- 
sinusoidal,  in  which  cases  the  relation  between  maximum,  mean 
and  virtual  values  is  quite  different. 


MEASUREMENT   OF   CURRENTS  AND   PRESSURES. 

The  dynamometer.  Consider  two  coils  of  wire  the  one 
carrying  the  steady  current  C^  the  other  the  steady  current  C2. 
Then  in  general  there  will  be  mechanical  forces  and  couples 
exerted  between  the  coils.  The  current  C^  will  produce  a  definite 
field  at  every  point  of  the  coil  carrying  C^,  which  field  will  be 
proportional  to  C\.  Since  the  force  between  any  element  of  the 


Fig.  4. 

coil  carrying  (72  and  the  field  produced  by  Cl  at  that  place  is 
proportional  to  the  product  of  the  current  (72  into  the  field  in 
which  it  is  placed,  the  force  or  couple  between  the  two  will  be 
proportional  to  the  product  (7,(72.  If  the  coils  be  symmetrical 
about  a  common  axis,  and  have  their  normals  at  right  angles,  it  is 


6  ALTERNATING    CURRENTS 

evident  that  from  symmetry,  only  a  couple  will  be  produced. 
Now  let  the  coil  carrying  (72  be  freely  suspended  and  provided 
with  some  means  for  allowing  the  current  to  flow  in  and  out  of  the 
coil  without  producing  any  friction,  such  as  having  its  ends  dipping 
in  cups  filled  with  mercury.  This  coil  will  tend  to  be  turned  by 
the  couple  into  such  a  position  that  the  flux  it  embraces  is  the 
maximum  possible.  If  by  any  means  we  apply  an  opposing 
couple,  capable  of  measurement,  of  amount  exactly  equal  to  that 
between  the  coils,  the  value  of  this  couple  will  measure  the 
product  of  the  currents. 

An  instrument  made  on  these  principles  is  called  an  electro- 
dynamometer  or,  for  shortness,  simply  a  dynamometer.  Fig.  4 
shows  such  an  instrument  of  a  usual  type.  In  this  form  the 
normal  configuration  of  the  system  is  such  that  the  axes  of  the 
two  coils  are  perpendicular,  and  this  condition  is  shown  to  exist 
by  the  pointer  /  which  is  attached  to  the  swinging  coil  A,  being 
midway  between  the  stops  S.  The  swinging  coil  is  carried  by 
a  fine  torsionless  thread  attached  at  the  bottom  to  the  coil  and  at 
the  top  to  a  pin  fixed  to  the  movable  torsion  head  H.  By  adjusting 
this  pin  the  coil  can  be  made  to  swing  quite  freely  with  its  ends 
dipping  into  the  two  mercury  cups  shown.  A  helical  spring  is 
attached  at  one  end  to  this  coil  and  at  the  other  end  to  the  torsion 
head,  as  shown,  so  that  the  indications  of  this  head  on  the  scale 
shown  will  measure  the  couple  that  is  being  mechanically  applied 
to  the  swinging  coil.  The  fixed  coil  B  is  attached  to  the  frame 
of  the  instrument,  and  the  two  coils  are  connected  to  the  sources 
of  the  currents  (7X  and  C2.  When  the  currents  pass,  the  torsion 
head  is  turned  till  the  index  /  is  midway  between  the  stops,  and 
if  a  is  the  nett  angle  of  twist  applied  as  read  by  the  pointer  P,  we 
must  then  have 

C,C2  =  k*.  a. 

For  the  couple  due  to  the  interaction  of  the  two  currents  is 
proportional  to  the  product  C^C*,  while  the  couple  due  to  winding 
up  a  helical  spring  is  proportional  to  the  angle  of  twist  imparted 
to  it. 

If  the  same  current,  (7,  be  sent  through  the  two  coils  put  in 
series  we  evidently  have  C  =  k^Ja. 

Now  let  the  current  be  an  alternating  one  of  any  form :  at  any 
instant  the  couple  will  then  be  proportional  to  the  square  of  the 
current  at  that  instant.  Owing  to  the  inertia  of  the  suspended 
coil  the  actual  mean  couple  experienced  by  the  suspended  coil  will 
be  the  mean  of  the  instantaneous  couples,  in  other  words  when  the 
torsion  head  is  turned  till  the  initial  configuration  is  reproduced,  the 
reading  of  the  head  will  be  a  measure  of  the  mean  of  the  instan- 
taneous squares  of  the  current,  or  if  ^  is  the  square  root  of  this 
ssum,  that  is  the  virtual  current,  we  have  ^  =  k\/a.  It  should  be 
noted  that  the  value  of  the  multiplier  k  is  the  same  as  when  a 


SIMPLE   HARMONIC   QUANTITIES  7 

steady  current  is  flowing.  The  dynamometer  has,  then,  the  very 
important  property  that  the  constant  obtained  in  calibration  with 
steady  currents  can  be  used  for  measuring  the  virtual  value  of  an 
alternating  one  independent  of  the  periodicity  or  form  of  that 
current.  There  are  many  other  current  measuring  instruments  or 
ammeters  which  can  be  used  for  the  measurement  of  this  quantity, 
but  few  of  them  possess  this  valuable  property. 

For  reasons  that  will  be  better  appreciated  later  on  it  is  very 
difficult  to  make  a  pressure  measuring  instrument  of  this  type  in 
the  ordinary  way,  that  is  by  giving  the  circuit  of  the  instrument  a 
very  high  resistance. 

Hot-wire  Instruments.  Another  property  depending  on  the 
square  of  the  current,  and  thus  capable  of  being  used  for  the 
measurement  of  a  virtual  current,  is  the  heating  of  a  wire.  If 
a  current  be  flowing  down  a  wire  the  rate  of  production  of  heat  is 
proportional  to  the  square  of  the  current,  and  hence  the  mean  rate 
of  production  will  be  proportional  to  the  square  of  the  virtual 
value  of  that  current  if  it  be  alternating.  A  wire  carrying  such 
a  current  will  eventually  get  to  a  steady  temperature  which  will 
depend  on  the  emissivity  of  the  wire  and  the  mean  rate  of 
production  of  heat.  In  consequence  of  this  rise  of  temperature 
it  will  increase  in  length,  and  the  increase  will  be  a  measure  of  the 
square  of  the  virtual  current.  This  alteration  in  length  is  as 
a  rule  very  small  and  a  common  way  of  magnifying  it  is  shown  in 
Fig.  5. 

The  wire  W,  W  is  attached  to  a  base  which  has  the  same 
temperature  coefficient  as  the  wire  so  that  any  alteration  of 
temperature  of  the  whole  instrument  will  not  affect  the  length 


of  the  wire.  A  small  sag  is  allowed  in  the  wire  which  is  taken  up 
by  a  fine  wire  which  is  fixed  at  one  end  F  to  the  frame  of  the 
instrument,  and  pulls  sideways  at  the  point  M  as  shown.  This 
wire  in  turn  has  a  slight  sag  which  is  taken  up  by  a  fine  thread 
attached  to"  it  at  N,  the  other  end  0  being  attached  to  a  spring  S  fixed 
to  the  case,  on  its  way  this  thread  passes  round  a  pulley  P  which  has 


8  ALTERNATING   CURRENTS 

attached  to  it  the  pointer  of  the  ammeter  and  is  pivoted  to  the  frame. 
It  will  be  seen  that  any  sag  in  the  original  wire  is  thus  greatly 
magnified.  It  will  be  evident  from  the  method  of  operation  of 
the  ammeter  that  if  the  instrument  be  calibrated  with  steady 
currents  the  same  calibration  is  correct  for  measuring  virtual 
currents.  The  wire  is  in  general  very  fine  and  only  capable  of 
carrying  very  small  currents.  When  it  is  required  to  measure 
currents  of  ordinary  magnitude  this  fine  wire  is  shunted  with  a 
suitable  shunt.  By  the  use  of  proper  precautions  it  is  possible 
to  use  this  type  of  instrument  for  measuring  virtual  volts ;  for 
this  purpose  a  high  series  resistance  is  used  just  as  in  the  case  of 
an  ordinary  volt  meter,  but  this  resistance  must  possess  the 
property  of  being  quite  "  non-inductive  "  for  a  reason  that  we  shall 
see  later  on. 

Electrostatic  voltmeter.  It  is  possible  to  use  the  properties 
of  condensers  for  the  purpose  of  measuring  virtual  pressures.  We 
know  that  the  energy  stored  in  a  condenser  is  given  by  the 
expression  ^e2.F,  where  e  is  the  applied  pressure,  and  F  the 
capacity  of  the  condenser. 

Let  a  condenser  be  provided  with  one  of  its  plates  capable  of 
rotation  about  an  axis  :  one  form  is  shown  diagram  matically  in 
Fig.  6.  Then  if  by  any  means  the  relative  position  of  the 
condenser's  plates  changes,  the  capacity  of  the  whole  will  change 


Fig.  6. 

by  a  definite  amount.  The  change  of  capacity  will  depend  on  the 
form  of  the  plates  and  the  angle  through  which  the  movable  one 
turns,  and  for  a  definite  form  of  plate,  it  will  depend  on  the  angle 
only.  Thus  if  the  plate  move  through  an  angle  a.  the  capacity  will 
change  by  some  definite  amount,  /(a).  If  a  pressure  e  is  acting 
between  the  plates  the  consequent  change  of  energy  will  be 
^e2y*(a).  Now  if  the  moving  plate  be  provided  with  a  controlling 
couple,  any  motion  will  produce  a  couple  tending  to  turn  the  plate 
back,  and  this  couple  will  depend  solely  on  the  angle  of  rotation 


SIMPLE   HARMONIC   QUANTITIES  9 

when  the  nature  of  the  control  is  fixed.  Thus  again  the  work 
done,  being  dependent  on  the  integral  of  the  product  of  the  couple 
and  the  corresponding  small  angular  rotation,  will  be  a  function  of 
that  angle  only;  let  it  be  denoted  by  F  (a).  When  the  plate  gets 
to  a  position  of  equilibrum,  the  two  amounts  of  work  must  be 
equal,  which  gives 


or 


Hence  there  will  be  a  definite  relation  between  the  pressure 
and  the  resulting  angular  motion  of  the  pivoted  plate,  and  it 
follows  that  this  angle  can  be  taken  as  a  measure  of  the  pressure. 

When  alternating  pressures  are  employed,  it  will  follow  in 
exactly  the  same  way  as  in  the  case  of  the  dynamometer,  that  the 
mean  position  of  the  plate  will  correspond  to  the  mean  of  the  squares 
of  the  instantaneous  pressures,  since  the  inertia  of  the  suspended 
plate  will  integrate  up  all  the  instantaneous  applied  couples.  Thus 
the  instrument  being  calibrated  with  steady  pressures  will  indicate 
on  the  same  scale  the  value  of  the  virtual  pressure  when  alternating 
pressures  are  employed.  Such  an  instrument  is  called  an  electro- 
static voltmeter. 

Periodicity  measurement.  It  is  very  often  of  importance 
in  laboratory  work  to  have  some  means  of  ascertaining  the 
periodicity  of  the  current  that  is  being  used.  This  is  usually 
found  by  means  of  the  forced  oscillations  of  a  tuned  reed.  If  a 
reed  of  steel  be  placed  in  the  field  of  an  electromagnet  which  is 
being  energised  by  the  alternating  current  it  will  vibrate  very 
strongly  when  the  natural  period  of  the  reed  is  the  same  as  that 
of  the  alternating  magnetic  field  acting  on  it.  Two  methods  can 
be  adopted  ;  the  more  accurate  is  to  provide  a  set  of  tuned  reeds 
whose  periodic  times  are  within  the  desired  range  of  periods  and 
which  differ  successively  by  say  two  periods  per  second  in  their 
own  pitch.  By  presenting  these  reeds  in  succession  to  the  magnet 
the  periodicity  of  the  current  can  be  found  from  noting  the  reed  or 
pair  of  reeds  that  respond.  A  single  reed  could  be  used  if  it  was 
possible  to  arrange  so  that  its  natural  period  could  be  conveniently 
altered.  Such  alteration  is  readily  produced  by  changing  the  length 
of  the  reed,  and  the  periodicity  teller  due  to  Mr  Campbell  acts  in 
this  way.  A  reed  is  taken  whose  length  can  be  altered  at  will  by 
means  of  a  rack  and  pinion  arrangement,  the  pinion  of  which  is  also 
attached  to  a  pointer  working  over  a  graduated  circular  scale.  The 
reed  is  acted  on  as  before  mentioned  by  an  electromagnet  which  is 
generally  placed  with  its  winding  in  circuit  with  an  incandescent 
lamp  suited  to  the  supply  pressure,  and  the  condition  of  synchro- 
nism between  the  reed's  period  and  that  of  the  current  is  shown  by 
the  sounding  of  the  reed  when  it  is  caused  to  vibrate.  The  calibra- 
tion of  the  instrument  is  best  performed  experimentally. 


10 


ALTERNATING   CURRENTS 


In  cases  where  a  small  range  of  periodicity  only  is  required, 
the  following  instrument,  due  to  Mr  Frahm,  is  convenient.  Con- 
sider a  set  of  little  reeds  all  exactly  tuned  to  vibrate  at  known 
periods  differing  say  by  half  a  period  from  one  another  and 
covering  the  range  of  periods  required.  Let  these  all  be  fixed  to 
a  base  forming  a  sort  of  comb.  Then  if  this  comb  be  shaken,  the 
reed  which  has  a  natural  period  equal  to  that  of  the  shaking  period 
will  be  violently  agitated,  the  others  being  practically  at  rest. 
Such  a  comb  of  reeds  is  arranged  in  such  a  way  that  it  can  be 
acted  on  by  an  electromagnet  attracting  the  base  to  which  the 
comb  is  fixed,  and  the  magnet  is  supplied  with  a  current  from  the 
source  of  energy  of  which  the  periodicity  is  to  be  found.  The 
required  periodicity  will  be  shown  by  the  corresponding  reed 
vibrating :  if  the  period  lies  between  those  of  two  adjacent  reeds, 
each  will  respond,  but  to  a  diminished  extent.  Thus  the  periods 
can  be  found  with  quite  sufficient  accuracy  for  most  purposes. 
Such  a  set  of  reeds  has  the  advantage  of  permanence. 

Angle  of  Lag  or  Lead.  In  many  cases  we  have  to  consider 
problems  where  several  simple  harmonic  quantities  all  having  the 
same  periodic  time  are  existing  at  the  same  time.  For  the  sake 
of  simplicity  consider  two  only  having  the  amplitudes  A  and  B. 
If  these  two  attain  their  zero  values  at  the  same  instant,  as  shown 
in  Fig.  7,  the  two  are  said  to  be  in  phase  and  will  be  represented  by 


Fig.  7. 

two  vectors  drawn  in  the  same  direction  but  of  different  lengths. 
It  may  happen  that  the  zero  of  one  is  not  attained  till  some  definite 
fraction  of  an  alternation  after  the  other  attains  its  zero  in  which 
case  there  is  said  to  be  a  difference  of  phase  between  the  two. 
If  they  attain  those  values  as  shown  in  Fig.  8  they  are  said  to 


SIMPLE    HARMONIC   QUANTITIES 


11 


be  antiphased  and  will  be  represented  by  two  vectors  drawn  in 
opposite  directions.     Again   let  the  two  quantities  be  given  by 


B 


Fig.  8. 


Fig.  9. 


the  curves  in  Fig.  9.  One  of  the  quantities  attains  its  zero  at  the 
value  of  the  angle  given  by  the  origin,  that  is  the  arbitrarily  chosen 
initial  zero  value,  while  the  other  does  not  attain  its  zero  till  the 
angle  X  has  been  traversed ;  hence  the  angle  X  will  be  the  difference 
of  phase  between  the  two.  In  the  figure  it  will  be  seen  that  B  does 
not  attain  its  zero  till  after  A  has  done  so,  and  this  is  expressed  by 
saying  that  B  lags  after  A.  On  the  other  hand,  if  we  are  referring 
matters  to  the  curve  B  it  is  seen  that  A  attains  its  zero  before  B, 
which  is  expressed  by  saying  that  A  leads  B.  The  vector  repre- 
sentation of  this  case  is  in  the  figure ;  the  two  vectors  are  drawn 
of  the  lengths  corresponding  to  A  and  B  and  with  the  angle 
X  between  them.  Since  positive  time  has  been  taken  to  coincide 
with  counter-clockwise  revolution,  the  angle  X  must  be  taken  as  in 
the  figure,  in  order  that  B  may  have  its  maximum  projection  after 
A.  When  the  angle  of  lag  or  lead  becomes  a  right  angle,  the  two 
quantities  are  said  to  be  "in  quadrature  "  or  sometimes  "at  quarter 
point."  In  this  case  if  one  be  given  by  a  =  A  sin  6,  the  other  can 
be  written  b  =  B  cos  6. 

With  more  than  two  such  vectors  representing  other  simple 
harmonic  quantities  each  must  be  drawn  with  its  proper  phase 
angle  with  respect  to  one  selected  vector. 

The  analytical  representation  of  an  angle  of  lag  or  lead  can 
be  easily  derived.  Let  the  quantity  a  be  given  by 

a  =  A  sin  6, 

then  the  other  quantity  will  be  expressed  by 
b  =  B  sin  (6  -  X). 

For  the  angle  B  being  reckoned  from  the  origin  as  shown,  b  does 
not  attain  its  zero  value  till  the  angle  X  has  been  traversed,  hence 
the  value  of  6  at  which  6  is  zero  is  X.  It  follows  that  the  expression 


12  ALTERNATING   CURRENTS 

above  satisfies  this  condition,  for  when  6  is  equal  to  X,  sin(#  —  \) 
is  zero.  Thus  a  negative  sign  to  the  phase  angle  implies  that  the 
corresponding  quantity  is  lagging  on  the  standard  quantity. 

It  will  readily  be  seen  that  the  converse  holds,  that  is  to  say 
a  positive  sign  to  X  means  a  lead.  For  since  b  lags  after  a  it 
follows  that  a  leads  6.  Now  instead  of  taking  a  as  the  standard 
let  us  take  6,  then  it  must  be  written  as  6  =  B  sin  6  and  the  origin 
is  now  the  point  where  b  is  zero.  But  a  has  attained  the  zero 
value  before  6  has  done  so,  and  at  the  time  when  B  was  (—  X). 
It  evidently  follows  that  the  expression  for  a  will  now  be 

a  =  A  sin  (0  +  X). 

Thus  the  question  of  lead  or  lag  and  its  mathematical  expression 
must  depend  on  which  of  the  quantities  is  taken  as  the  standard 
of  reference. 

It  is  evident  that  with  these  sinusoidal  quantities  any  two 
symmetrically  situated  points,  as  for  example  the  maxima,  could 
have  been  taken  in  considering  the  question  of  relative  phase. 

Summation  and  resolution  of  harmonic  quantities.    The 

summation  of  simple  harmonic  quantities  can  be  made  by  the 
ordinary  parallelogram  method,  but  the  quantities  must  necessarily 


B 

Fig.  10. 

be  of  the  same  nature.  Thus  in  Fig.  10  if  OA  and  OB  be  two  such 
quantities,  the  resultant  will  be  given  by  the  diagonal  OR. 
Similarly  the  difference  will  be  given  by  the  other  diagonal  AB, 
the  direction  of  this  diagonal  will  depend  on  which  vector  is  being 
subtracted ;  if  OA  is  taken  from  OB  the  arrow-head  must  be  put 
pointing  from  A  to  B,  while  if  OB  is  taken  from  OA.  it  must  point 
from  B  to  A.  If  necessary  this  difference  vector  can  be  drawn 
from  the  origin  in  its  proper  direction. 

It  also  follows  that  any  given  quantity  can  be  resolved  into  two 
components  along  any  desired  directions,  the  most  useful  directions 
are  in  general  perpendicular  to  each  other.  Thus  in  Fig.  11  let  OA 
represent  one  quantity  and  OB  another  of  a  different  class,  lagging 
by  X  on  OA .  Then  we  can  resolve  OB  into  two  components,  for 


SIMPLE    HARMONIC    QUANTITIES 


13 


example  one  along  OA  called  the  in-phase  component  0(7,  and  the 
other  at  right  angles  thereto,  called  the  quadrature  component  CB. 
The  maximum  values  of  these  components  are  evidently  given  by 
B  cos  X  for  the  in-phase  one,  and  B  sin  X  for  the  other,  and  the 
latter  lags  a  right  angle  on  the  former.  The  corresponding  repre- 
sentation by  curves  can  be  found  thus  :— 

If  OA  is  as  before  a  =  A  sin  0  then  OB  is  6  =  B  sin  (6  -  X).  The 
latter  can  be  written  6  =  (B  cos  X)  sin  6  —  (B  sin  X)  cos  6,  showing 
that  it  consists  of  the  simple  harmonic  of  maximum  value  B  cos  X, 
which  is  in  phase  with  OA  and  the  simple  harmonic  of  maximum 
value  B  sin  X  which  is  in  quadrature  with  it.  The  curves  repre- 
sented by  these  expressions  are  shown  in  Fig.  11,  the  two  components 


Fig.  11. 

of  b  being  dotted  in.  This  resolution  into  the  two  perpendicular 
components  is  very  important,  and  will  be  often  required.  Mani- 
festly any  number  of  vectors  of  the  class  OB  can  be  similarly 
resolved  in  the  direction  of  OA  and  at  right  angles,  and  the  sum 
of  the  separate  projections  will  thus  give  the  total  components  in 
the  two  given  directions. 

The  rate  of  change  and  integral  of  a  simple  harmonic 
quantity.  In  many  problems  the  relation  between  a  simple 
harmonic  quantity  and  its  rate  of  change  or  its  integral  is  of  very 


14 


ALTERNATING   CURRENTS 


great  importance.     Let  us  consider  the  case  where  it  is  expressed 
in  terms  of  the  time,  then 


Thus  we  have      -j-  = 


A  sin  pt 
cos  pt 


and  also 


fadt  = cospt 


-  or  - 


a  =  A  cos  pt. 

da  . 


fadt  =  —  sinpt. 


In  the  latter  case  the  constant  of  integration  has  been  taken 
as  zero,  which  is  always  the  case  in  alternate  current  work.  These 
results  will  be  seen  to  be  represented  by  the  curves  in  Fig.  12 


PA 


Fig.  12. 

where  the  axes  for  the  sine  case  and  for  the  cosine  case  are  indicated. 
In  both  it  will  be  noticed  that  the  rate  of  change  leads  on  a  by 
a  right  angle  while  the  integral  of  a  lags  a  right  angle,  further  the 
former  is  p  times  as  great  as  a  while  the  latter  is  1/pth  of  a.  The 
vector  representation  of  these  cases  is  evidently  as  shown  in  the 
same  figure. 


CHAPTER   II. 

CURRENTS  DUE   TO   SIMPLE   HARMONIC   PRESSURES. 

Current  due  to  Simple  Harmonic  E.M.F.s.  If  a  simple 
harmonic  E.M.F.  be  applied  to  a  circuit  containing  nothing  but 
ohrnic  resistance  it  is  evident  that  at  every  instant  this  E.M.F.  has 
only  to  overcome  the  resistance  of  the  circuit ;  in  other  words  if  c 
denote  the  resulting  current  at  any  moment,  e  the  corresponding 
E.M.F.  and  R  the  resistance,  we  must  have  e  =  cR,  thus  the  E.M.F. 
given  by  the  equation  e=Esinjt^  will  produce  the  current 

c  =  -^  sin  pt.     In  such  a  case  the  current  curve  will  be  an  exact 

copy  of  the  E.M.F.  curve  and  there  will  be  neither  lead  nor  lag. 
It  will  be  seen,  further,  that  even  if  the  E.M.F.  be  non-sinusoidal 
the  same  relation  holds,  and  that  the  two  curves  will  have  exactly 
the  same  shape ;  this  point  will  be  of  importance  later  on. 

Circuit  with  Self-Induction.  If  a  circuit  contain  self- 
induction  as  well  as  ohmic  resistance  the  case  is  different.  At 
any  instant  the  current  produced  will  have  a  definite  rate  of 
change,  and  we  know  that  a  circuit  of  which  the  coefficient  of 
self-induction  is  L  is  such,  that  when  a  current  c  is  flowing  there 
is  a  flux  of  magnetism  of  the  amount  Lc  passing  through  it. 
Hence  there  will  be  an  E.M.F.  produced  in  the  circuit  itself,  and 
the  amount  of  that  E.M.F.  will  be  connected  with  the  rate  of  change 

dc 
of  the  flux  as  shown  by  the  equation  eg  =  —  L  -=- ,  eg  being  the 

instantaneous  value  of  this  induced  E.M.F.  The  ohmic  resistance 
of  the  circuit  will  demand  that  a  certain  pressure  be  supplied  to 
force  the  current  round  the  circuit,  and  the  amount  of  that 
pressure  will  be  er  =  cR.  Now  the  pressure  that  exists  at  any 
instant  at  the  terminals  of  the  circuit  must  be  of  such  an  amount 
as  to  just  suffice  to  send  the  current  down  the  resistance  and  to 
supply  a  pressure  equal  and  opposite  to  eg,  since  if  this  condition 
be  fulfilled  the  circuit  will  be  in  a  state  of  equilibrium.  Hence 


16  ALTERNATING   CURRENTS 

the  E.M.F.  e  applied  to  the  circuit  must  at  each  instant  be  given 
by  the  equation 

e  =  cR  4-  L  -j-  . 
at 

The  E.M.F.S  mentioned  above  receive  special  names,  e  is  called 
the  impressed  pressure,  es  the  induced  pressure,  its  negative  being 
known  as  the  back  or  reactance  pressure,  while  er  is  called  the 
effective  pressure.  It  should  be  noted  that  the  latter  is  the  only 
one  concerned  in  the  dissipation  of  energy,  the  term  due  to  the 
self-induction  corresponds  to  energy  which  is  alternately  stored  in 
and  restored  by  the  circuit. 

Impedance  and  Reactance.  Let  the  current  flowing  in 
the  above  circuit  be  taken  as  sinusoidal  and  be  represented  by 

c  =  C  sin  pt, 

dc 
so  that  -j-  =  pC  cos  pt. 

Then  the  equation  connecting  the  E.M.F.  and  this  current  will  be 

e  =  CR  .  sin  pt  +  pLC  cos  pt  =  C  (R  sin  pt  +  pL  .  cos  pt). 
If  we  put  I2  =  Rz  +  L'2p*, 

then  e  =  CI  (  -j  sin  pt  +  *y-  cos  pt  j  . 

This  equation  can  be  simplified  as  follows  :   let  X  be  the  angle 
given  by 


then         e  =  CI  (cos  X  .  sin  pt  +  sin  X  .  cospt)  =  CI  sin  (pt  +  X). 

If  we  denote  the  maximum  value  of  e  by>  E  we  have  E  =  C  .  /. 
That  is,  if  the  current  be  c  =  C  sin  pt  the  E.M.F.  will  be 

e  =  E  sin  (pt  +  X), 
the  values  of  E,  X  and  /  being  defined  as  above. 

If  instead  of  starting  with  the  current  as  given  we  take  the 
E.M.F.,  it  evidently  follows  that  when  the  E.M.F.  e  =  E  sin  pt  is 
applied  to  the  circuit  the  current  produced  will  be 

c  =  -j  sin  (pt  —  X). 

Hence  in  such  an  inductive  circuit  the  current  will  lag  after  the 
pressure  by   the  angle  whose  tangent  is  given  above,  and  the 

maximum  value  of  the  current  is  -j  instead  of  -~  as  would  be  the 
case  with  a  non-inductive  circuit.     The  quantity  I  is  called  the 


SIMPLE   HARMONIC   CURRENTS  17 

Impedance  of  the  given  circuit,  the  product  Lp  being  designated 
the  Reactance. 

The  following  case  is  of  interest  as  illustrating  the  above 
points.  Let  a  coil  be  made  of  a  large  number  of  turns  and  as  low 
a  resistance  as  possible  and  place  in  series  with  it  a  non-inductive 
resistance  such  as  one  or  more  incandescent  lamps.  If  the  pressure 
at  the  terminals  of  each  of  the  two  sections  and  at  the  terminals 
of  the  whole  be  measured,  it  will  be  found  that  the  square  of  the 
latter  is  very  approximately  equal  to  the  sum  of  the  squares  of  the 
former.  In  this  circuit  the  inductive  coil  may  be  looked  on  as 
producing  the  back  E.M.F.  in  the  whole  circuit,  while  the  lamps 
provide  its  resistance ;  we  can  thus  measure  separately  these 
components  which  it  is  not  usually  possible  to  do  in  an  ordinary 
circuit.  It  will  be  noted  that  complete  separation  of  the  two 
E.M.F.S  is  impossible  since  the  coil  must  dissipate  some  energy 
owing  to  the  impossibility  of  making  it  quite  devoid  of  resistance. 

Circuit  with  Capacity.  Consider  the  case  of  a  circuit  made 
up  of  a  resistance  in  series  with  a  condenser  of  capacity  F  and  let 
a  sine  E.M.F.  be  applied.  We  know  that  when  a  condenser  of 
capacity  F  has  a  pressure  of  e  volts  applied  to  its  terminals,  a 
quantity  of  electricity  given  by  q  =  eF  passes.  Hence  if  a  varying 
current  c  be  flowing,  since  c  =  dq/dt  the  relation  between  the 

pressure  and  this  current  is  c  —  F-r  or  the  pressure  is  related  to 

the  current  by  the  equation  e—  j^lc  .dt.  No  constant  of  integra- 
tion will  be  required  since,  in  the  case  of  an  alternating  pressure, 
no  permanent  charge  of  the  condenser  can  ensue.  If  the  current 
flowing  be  written  c  =  Csin^,  the  pressure  at  the  terminals  of 
the  whole  circuit  will  have  to  equilibrate  this  pressure  as  well  as 
to  supply  that  required  to  overcome  the  ohmic  resistance ;  it  will 

therefore  be 

/-\ 

e  =  CR  sin  pt  —  -~-  cos  pt. 
Hence  if,  as  in  the  last  case,  we  put  I2  =  Rz  +  -=^  --  ,  E  =  C  .  /,  and 

tan  X  =  -,      ,  we  can  easily  see  that  e=E.  sin  (pt  —  X). 

It  follows  that  if  the  pressure  be  given  by  e  =  E  sin  pt,  the 
current  will  be  given  by 

c  =  j  sin  (pt  +  X). 

Hence  the  effect  of  such  a  condenser  is  to  cause  the  current  to 
lead  the  pressure  at  the  terminals  of  the  circuit  by  an  angle 
whose  cotangent  is  FRp,  the  value  of  the  current  being  still 
diminished  in  this  case. 


18  ALTERNATING   CURRENTS 

The  case  where  all  three  quantities  R,  L  and  F  are  present  in 
series  in  the  same  circuit  can  be  readily  deduced.  It  can  be  seen 
at  once  that  if  the  applied  pressure  in  this  case  be  e  =  E  sin  pt, 
the  current  will  be 


c  =  -   sin  (pt  —  \), 


where 


and  tan  X 


~ 

=  -  ^  —  -  . 


Hence  we  may  have  either  an  angle  of  lag  or  one  of  lead  depend- 
ing on  whether  Lp  is  greater  or  less  than  l/Fp.  It  should  be 
noticed  that  if  LFp2  —  1  the  angle  of  phase  difference  vanishes, 
and  in  this  case  the  impedance  becomes  simply  equal  to  the 
resistance.  This  is  sometimes  referred  to  as  the  case  of  resonance, 
for  if  such  a  circuit  contained  capacity  and  self-induction  only,  the 
current  would  be  in  that  case  infinite.  Even  when  resistance  is 
present  this  relation  leads  to  the  production  of  very  high  pressures 
at  the  terminals  of  the  different  parts  of  the  circuit. 

Capacity  and  Impedance  in  Parallel.  The  following  case 
is  of  interest.  Let  an  inductive  circuit  be  in  parallel  with  a 
capacity.  We  will  show  that  for  some  definite  capacity  the 
current  is  a  minimum  and  then  the  pressure  at  the  terminals  and 
the  total  current  flowing  up  to  the  two  are  in  phase.  If  the 
letters  have  the  same  meaning  as  in  the  last  case,  and  if  cx  denote 
the  current  in  the  coil  and  c2  that  in  the  capacity  we  have 

cx  =  —  sin  (pt  —  X), 

de 
and  since  c2  =  F  -r  and  e  —  E  sin  pt  we  have 

c2  =  E.Fp  cos  pt. 
Hence  the  current  flowing  into  the  two  will  be  c  =  c±  +  c2  or 


c  =  -    {sin  (pt  —  X)  +  FIp .  cos  pt}. 


This  can  be  written 

E 
c 


—-f  (sin  pt  •  cos  \  —  cos  pt .  sin  X  -\-  FIp  .  cospt). 
But  sin  X  =  -*-   and  cos  X  =  -j . 

Hence  c=E\j2smpt  -  (-j^- -  Fpjcosptl . 


SIMPLE   HARMONIC  CURRENTS  19 

If  we  put  tan^  =  (L~J/2)P, 

(fLj)         \2     j?2)  i 
this  becomes       c  =  E  j(~77~^jp)  +  74f  sin(;rf  — ^). 

The  current  taken  by  the  two  in  parallel  will  evidently  be  a 
minimum  when  the  multiplier  is  a  minimum  and  since  it  is  a  sum 

of  two  squares  this  will  be  the  case  when  -~  =  Fp  or  L  =  FI-. 

Then  it  follows  that  tan  -\fr  is  zero,  or  the  current  and  the  terminal 
pressure  are  in  phase.  It  may  be  noted  that  when  the  resistance 
of  the  first  circuit  is  small  compared  with  the  self-induction, 
/  becomes  very  nearly  Lp,  and  hence  the  relation  giving  no  lead 
or  lag  is  LFpz  =1  or  , the  same  as  for  the  case  where  the  capacity 
and  induction  are  in  series. 

Vector  representations.  The  consideration  of  different 
forms  of  circuits  is  more  simply  carried  out  by  the  use  of  the 
vectorial  representation  and  we  will  now  briefly  work  out  a  few 
such  cases.  Consider  that  of  a  coil  possessing  resistance  and 
self-induction  only,  as  taken  on  p.  15.  It  was  there  shown 
that  such  a  circuit  possessed  a  quality  called  its  impedance,  and 
we  can  find  the  value  of  the  impedance  as  follows.  Draw  a  line, 
OR,  of  such  a  length  as  to  give  (on  an  appropriate  scale)  the 


Fig.  13. 

value  of  the  ohmic  resistance  of  the  circuit ;  the  product  Lp  for 
the  circuit,  or  its  Reactance  being  given,  let  a  perpendicular 
line,  LR,  be  drawn  to  give  the  value  of  this  quantity  on  the  same 
scale,  then  since  the  impedance  is  given  by  P=R2+L2p*,  it  is  evident 
that  /  will  be  represented  on  the  same  scale  by  OL ;  this  triangle 
is  called  the  impedance  triangle  for  the  circuit.  Now  the  angle 
LOR  is  such  that  its  tangent  is  Lp/R,  and  we  saw  that  in  the 
circuit  considered  the  current  lagged  after  the  pressure  by  that 
angle.  Thus  if  QV  represents  the  pressure,  the  current  flowing 
will  be  given  by  the  line  QC  drawn  at  the  angle  \  to  Q  V.  These 
two  lines  can  be  taken  as  giving  either  the  maximum  or  the 
virtual  value  of  the  corresponding  quantities  on  scales  appropriate 
to  either.  It  will  be  seen  that  if  QV  is  parallel  to  OL,  then  QC 
is  parallel  to  OR,  hence  if  OL  be  taken  as  the  direction  of  the 
pressure  the  line  OR  gives  the  direction  of  the  current. 

2—2 


20 


ALTERNATING   CURRENTS 


It  follows  that  we  can  adjust  the  scales  of  the  different 
quantities  in  such  a  way  that  OL  shall  either  represent  the 
maximum  pressure  in  the  circuit  or  the  virtual  value,  which  is 
taken  being  a  matter  of  convenience. 

In  this  case,  on  a  chosen  scale  of  pressure  such  that  OL  gives 
the  applied  pressure,  the  lines  OR  and  RL  must  also  represent 
pressures,  that  is  the  applied  pressure  can  be  considered  to  have 
those  components.  Now  if  C  be  the  current  flowing  it  is  evident 
that  OL  will,  on  the  scale  of  pressures,  be  of  the  length  C .  /,  and 
hence  the  pressure  denoted  by  OR  will  have  the  value  C .  R  and 
that  denoted  by  LR  will  have  the  value  Lp .  C;  the  former  pressure 
is  (as  before  mentioned)  the  effective  pressure  and  the  latter  the 
reactance  pressure,  the  first  represents  that  part  of  the  applied 
pressure  that  is  in  phase  with  the  current  and  is  operative  in 
forcing  the  current  against  the  resistance, '  and  the  latter  com- 
ponent is  operative  in  equilibrating  the  pressure  produced  by  the 
self-induction.  Thus  in  Fig.  14,  the  applied  pressure  OF  is 


Fig.  14. 

equivalent  to  the  perpendicular  components  OT  and  TV,  the 
pressure  induced  in  the  coil  due  to  its  self-induction  will  have 
the  direction  OS,  lagging  by  a  right  angle  on  OC.  This  lag  is 
due  to  the  fact  that  the  induced  E.M.F.  is  the  negative  change-rate 
of  the  flux,  and  since  on  p.  14  we  saw  that  the  change-rate 
of  OC  will  be  represented  by  the  vector  OU  in  the  figure,  its 
negative  must  be  represented  by  the  equal  and  opposite  line  OS. 
Hence  the  component  of  the  impressed  pressure  that  has  to 
equilibrate  the  self-induction  E.M.F.,  or  OS,  must  be  equal  to  OS 
and  be  in  the  opposite  direction  as  is  TV. 

It  will  be  seen  that  if  C  is  the  maximum  current,  the  maximum 
applied  pressure  is  C.I,  the  maximum  effective  pressure  is  C .  R 
and  the  maximum  back  or  reactance  pressure  is  Lp  .  C  ;  the  virtual 
values  of  the  same  quantities  will  be  1/V2  times  these  values  since 


OF 


SIMPLE   HARMONIC    CURRENTS 


21 


sinusoidal  variations  have  been  assumed  in  taking  the  vectorial 
representations  of  the  same. 

Since  the  reactance,  Lp,  always  occurs  as  a  single  quantity 
when  the  applied  pressure  has  definite  periodicity,  it  will  be 
denoted  by  the  symbol  S,  that  is  the  reactance  of  the  circuit  will 
be  S  where  S  =  Lp. 

Reactive  circuits.  Consider  as  an  example  the  case  where 
a  coil  has  a  resistance  of  10  ohms,  a  self-induction  of  0*03  and  the 
periods  are  such  that  2?m  is  500.  Then  R  =  10,  S  =  15  and  /  is 
nearly  18.  The  impedance  triangle  is  shown  in  Fig.  15.  It 


L,R. 


75 


10 


Fig.  15. 


follows  that  X  will  be  such  that  its  tangent  is  1*5  or  is  about 
56°  20',  so  that  whatever  pressure  is  applied,  the  angle  of  lag 
between  the  pressure  and  the  current  will  be  of  that  amount. 
Let  a  pressure  of  the  virtual  value  100  volts  be  used,  then  the 
current  will  be  100/18  or  5'55  amperes,  the  virtual  effective 
pressure  will  be  55'5  volts  and  the  virtual  back  pressure  will  be 
83'2  volts,  the  corresponding  maximum  values  being  \/2  times  as 
great. 

A  difficulty  is  sometimes  felt  in  the  case  where  the  resistance 
becomes  vanishingly  small  relative  to  the  reactance.  In  this  case 
the  line  OR  vanishes  but  the  current  still  flows  in  the  direction  of 
that  line,  that  is  in  the  limit,  at  right  angles  to  the  pressure. 
The  impressed  pressure  vector  and  the  back  pressure  vector  in 
that  case  exactly  oppose,  each  having  the  value  SC. 


*~ 23-5  V  - -65  V-~ 


Let  the  same  coil  be  connected  in  series  with  a  non-inductive 
resistance  of  5  ohms  as  shown  in  Fig.  16.     The  total  resistance 


22 


ALTERNATING   CURRENTS 


of  the  circuit  is  now  15  ohms  and  the  reactance  is,  as  before,  15, 
hence  the  total  impedance  is  21*2  ohms,  that  of  the  inductive  part 
itself  being  as  before  18.  If  a  terminal  pressure  of  100  volts  be 
applied,  the  current  will  be  100/21'2  or  47  amperes,  and  the  angle 
of  lag,  A,  between  this  pressure  and  the  current  will  be  evidently  45°. 
Thus  the  pressure  across  the  ends  of  the  non-inductive  resistance 
is  5  x  47  or  23'5  volts,  while  that  across  the  inductive  coil  is 
18  x  47  or  85  volts;  the  angle  of  lag,  \,  between  this  pressure 
and  the  current  is  the  same  as  in  the  first  case. 

Now  let  the  second  coil  have  in  addition  a  self-induction  such 
that  at  the  given  periodicity  its  reactance  is  2.  Then  its  impedance 
will  be  nearly  5'48.  Thus  the  two  impedance  triangles  are  as 
shown  in  Fig.  17.  Being  in  series  the  whole  circuit  acts  as  if  it 


Fig.  17. 

had  a  resistance  of  15  ohms  and  a  reactance  of  17,  hence  the 
complete  figure  is  as  drawn.  The  total  impedance  is  227  ohms, 
and  hence  with  a  pressure  across  the  terminals  of  100  volts,  the 
current  taken  will  be  4*4  amperes  and  the  angle  of  lag,  X,  will  be 
such  that  its  tangent  is  17/15  or  will  be  about  45^°.  The  pressure 
across  the  first  coil  will  be  79'2  volts  and  that  across  the  other 
24'2  volts,  the  angle  of  lag,  \i,  for  the  first  being  still  56°  20'  while 
for  the  latter  the  angle,  A2>  is  22°.  Certain  arrangements  of  the 
circuits  may  result  in  the  sum  of  the  pressures  being  the  same  as 
the  applied  pressure,  for  example  if  the  two  impedance  triangles 
are  similar  this  is  evidently  the  case. 

Condenser  circuits.  The  vector  representation  of  the  case 
of  the  condenser  circuit  considered  on  p.  17  can  be  treated  in  the 
same  way.  The  vector  triangle  can  be  drawn  exactly  as  in  the 
last  case,  care  being  taken  however  to  note  that  the  current  leads 
the  pressure.  Thus  if  the  condenser  in  circuit  have  a  capacity  of 
25  microfarads  and  the  periods  multiplied  by  2?r  be  500,  the  value 
of  TL/Fp  will  be  80,  for  it  must  be  remembered  that  F  has  to 
be  measured  in  farads :  let  the  resistance  be  60  ohms,  then  the 
vector  triangle  will  be  as  in  Fig.  18,  the  current  vector  being  drawn 
with  a  negative  angle  relative  to  the  potential  vector  in  order  to 
show  that  the  phase  angle,  whose  tangent  is  \jFRp  must  be  taken 


SIMPLE   HARMONIC   CURRENTS 


23 


as  giving  the  current  a  lead  on  the  pressure  at  the  terminals. 
In  this  case  the  impedance  is  100,  so  that  if  a  pressure  of  1000 


60 


.  v.—  » 


80 


< -1OOO.V.—-* 


78 


60 


80 


1666.V. 


100 


60 


80 


Fig.  18. 


volts  be  maintained  across  the  whole  circuit,  the  current  will  be 
10  amperes,  the  effective  pressure  will  be  600  volts,  and  the 
pressure  on  the  condenser  will  be  800  volts.  The  angle  of  lag  has 
a  tangent  given  by  4/3  and  is  hence  about  53°. 

The  case  where  the  resistance  is  inductive  can  be  similarly 
treated  :  let  the  coil  have  a  reactance  of  50.  Draw  the  impedance 
triangle  for  the  coil  as  shown,  and  draw  backwards  from  the  top  angle 
a  line  equal  to  the  value  of  l/FRp,  that  is  to  80.  The  closing 
line  of  the  lower  triangle  evidently  gives  the  impedance  of  the 
whole  circuit.  Its  value  is  67,  and  the  current  with  a  terminal 
pressure  of  1000  volts  is  about  15  amperes,  hence  the  pressure  on 
the  ends  of  the  coil  is  1170  volts  while  that  across  the  condenser 
is  1200;  the  angle  of  lead  is  reduced  to  27°.  In  this  case,  then. 


24 


ALTERNATING   CURRENTS 


either  of  the  pressures  can  be  greater  than  that  across  the  whole 
system. 

Take  the  case  where  the  reactance  is  equal  to  the  condenser 
effect,  that  is  to  say  when  IjFp  is  equal  to  Lp  or  FLp*  =  1.  In 
this  case  the  two  vertical  lines  being  equal  annul  one  another,  and 
the  total  impedance  reduces  to  the  ohmic  resistance,  while  the 
coil's  impedance  is  100.  Thus  the  current  with  1000  volts 
terminal  pressure  will  be  16f  amperes,  the  pressure  on  the  coil 
will  be  1666  volts,  and  that  on  the  condenser  will  be  1333  volts. 
It  will  readily  be  seen  that  when  the  resistance  is  small,  the 
pressures  on  the  two  parts  of  the  circuit  may  be  many  times  the 
terminal  pressure. 

Circuits  in  parallel.  We  will  now  consider  the  case  of  two 
circuits  in  parallel,  and  as  an  example  take  the  case  where  the 
resistance  of  one  circuit  is  3  ohms  and  its  reactance  is  4  while  for 


Fig.  19. 

the  other  circuit  the  resistance  is  6  and  the  reactance  is  3.  The 
impedance  triangles  are  shown  in  Fig.  19  and  the  respective 
values  of  the  impedance  are  5  and  6'5.  Let  the  two  be  in  parallel 


SIMPLE    HARMONIC   CURRENTS  25 

and  have  an  applied  pressure  of  100  volts  across  their  common 
terminals.  Draw  a  semicircle  as  shown  with  a  diameter  equal  to 
the  value  of  the  applied  pressure  on  any  assigned  scale,  then  the 
current  taken  by  the  first  coil  will  be  20  amperes,  the  back 
pressure  will  be  80  and  the  effective  pressure  will  be  60,  the 
pressure  triangle  being  shown  with  its  sides  numbered.  Similarly 
for  the  second  coil  the  current  will  be  15'4  amperes,  the  back 
pressure  46'2  and  the  effective  92'4,  the  pressure  triangle  being 
also  shown.  Then  the  phases  of  the  currents  will  be  those  of  the 
effective  pressures ;  on  the  lines  representing  these  pressures  are 
set  off  to  any  desired  scale  of  current  the  values  of  the  currents  as 
shown  at  OCl  and  OC2.  The  resultant  current  will  then  be  given 
by  OC  and  its  phase  angle  by  X. 

The  value  of  the  current  will  on  scaling  off  be  found  to  be 
35  amperes.  Evidently  the  two  coils  can  then  be  replaced  by 
a  single  coil  of  such  constants  that  it  carries  the  current  of 
35  amperes,  and  its  impedance  triangle  is  as  shown,  the  sides 
being  65,  75  and  100.  Hence  the  back  pressure  for  that  equivalent 
coil  will  be  65  volts  and  the  effective  pressure  will  be  75  volts  as 
shown,  thus  the  tangent  of  X  is  65/75  or  about  0'88,  that  is  X 
is  about  41 1  °.  The  equivalent  resistance  will  be  found  by 
dividing  the  effective  pressure  by  the  current,  35  amperes,  and 
is  2*62  ohms,  similarly  the  equivalent  reactance  is  2'27. 

Instead  of  actually  drawing  in  the  current  vectors  the  value 
and  phase  angle  for  the  resultant  can  be  found  in  the  usual 
algebraic  manner.  Thus  if  ^  and  ^  be  the  virtual  component 
currents  and  X!  and  \^  their  phase  angles,  if  we  resolve  the 
currents  along  the  pressure  line  and  perpendicular  thereto  we 
evidently  get 

X  =  $;  cos  Xj  +  ^2  cos  X.2  =  X 


sin.  X  =  Wj  sm  Xj  +  w-i  sin  X,  =  F, 
which  lead  to      ^2  =  X2  +  F2  and  tan  X  =  F/X. 
In  the  present  case  these  equations  are 

X  =  (20  x  0-6)  +  (15-4  x  0-92)  =  26'2, 
X  =  (20  x  0-8)  +  (15-4  x  0-46)  =  23'2 ; 

the  sines  and  cosines  of  X!  and  X,  being  readily  found  from  the 
impedance  triangles. 

This  gives  ^=35  and  tan  X  =  0'88,  the  same  as  the  graphical 
solution. 

As  an  example  of  the  choice  of  a  special  scale  for  the  current 
vector  we  will  consider  the  case  of  the  condenser  in  parallel  with 
an  impedance  coil  (p.  18).  Let  OF,  Fig.  20,  be  the  impressed 
pressure  vector  and  let  OVR  be  the  usual  impedance  triangle  for 
the  coil  giving  its  effective  and  back  pressures.  The  line  OR 


26 


ALTERNATING  CURRENTS 


gives  the  direction  of  the  current  in  the  coil,  and  by  suitably 
selecting  the  scale  of  current,  it  can  be  likewise  taken  to  give  its 
magnitude.  The  current  flowing  in  the  condenser  will  have  the 
value  EFp  and  will  lead  the  pressure,  OF,  by  a  right  angle,  being 


given  by  the  line  OF.  Now  when  the  resulting  circuit  is  such 
that  the  phase  angle  vanishes,  the  current  flowing  into  the 
circuit  must  be  such  that  it  lies  along  OF,  hence  the  current 
in  the  coil  must  be  such  as  to  give  this  current  when  it  is 
combined  with  that  through  the  condenser.  Draw  the  line  RX 
perpendicular  to  OF,  then  the  coil-current  is  equivalent  to  the 
currents  OX  and  RX  of  which  the  former  is  in  phase  with  the 
pressure,  the  latter  with  the  condenser  current.  Hence  when  the 
line  current  is  in  phase  with  the  pressure,  it  will  be  given  by  OX 
and  the  other  component  of  the  coil  current  (that  is  RX)  must  be 
equal  to  the  condenser  current.  But  if  /  is  the  impedance  of  the 
coil  and  E  the  pressure,  the  coil  current  is  E//,  and  the  component 

RX  is  E  sin  X ;  but  sin  X  is  -y- ,  and  hence  the  component  RX  is 

•'    P .     But  this  is  equal  to  the  condenser  current  or  EFp,  arid 

hence  we  have 

E.Lp=EFp.I2  or  L  =  F.I\ 

the  result  proved  before. 

It  is  unnecessary  to  multiply  examples,  since  whether  the 
circuits  have  induction  or  capacity  the  same  construction  can  be 
applied,  but  further  examples  will  occur  incidentally  in  the  con- 
sideration of  various  problems. 


CHAPTER   III. 

MEASUREMENT   OF  POWER. 

Power.     Simple  harmonic  pressure  and  current.     We 

have  seen  that  in  general  the  current  produced  in  a  circuit  by  an 
alternating  pressure  is  out  of  phase  with  it.  The  rate  of  doing 
work  in  such  a  circuit  will  therefore  not  only  vary  from  instant  to 
instant  but  may  be  negative  at  certain  points  in  the  alternation. 
For  example  in  Figs.  21  and  22  are  drawn  two  curves,  the  one  of 
pressure  the  other  of  the  resulting  current.  A  third  curve  is 
drawn  such  that  at  each  point  its  ordinate  is  equal  to  the  in- 
stantaneous product  of  the  two  others,  in  other  words,  this 
represents  the  instantaneous  rate  of  working  in  the  circuit.  If 
the  original  curves  be  both  sine  curves  it  is  easy  to  evaluate  the 
instantaneous  rate  of  working.  For  if  the  pressure  be  given  by 
e  =  E  sin  6  and  the  current  produced  by  c  =  C  sin  (6  —  X)  the 
instantaneous  rate  of  working  is 

w  =  E  .  C  .  sin  6 .  sin  (0  -  X). 
This  can  be  written  in  the  form 

w  =  ^  {cos  X  -  cos  (2(9  -  X)}, 

which  shows  that  the  power  is  a  function  of  the  time  which  is  of 
one  half  the  periodic  time  of  the  components.  It  will  be  seen  that 
the  power  has  two  positive  portions  in  one  alternation,  and  two 
negative,  vanishing  therefore  four  times  per  alternation.  During 
the  positive  portions  work  is  being  done  by  the  pressure  on  the 
circuit,  and  during  the  others  the  energy  stored  in  electromagnetic 
or  condenser  action  is  given  back  to  the  generator.  When  the 
angle  of  phase  difference  is  zero  we  have 

EC* 

w  =  =£-  (1  -  cos  2(9), 

showing  that  there  is  no  negative  power,  but  that  it  falls  to  zero 
only  twice  in  an  alternation.  This  must  be  the  case  as  there  is  in 


28  ALTERNATING   CURRENTS 

this  condition  no  means  by  which  energy  can  be  stored.  The 
other  limiting  case  is  when  the  phase  angle  is  90°,  when 

w  =  -~-  sin  20, 

showing  that  the  positive  and  negative  regions  are  equal,  and 
hence  no  nett  work  is  done  in  the  circuit.  This  again  must  be 
the  case  since  the  phase  angle  being  90°  infers  that  the  resistance 
is  zero,  and  thus  energy  dissipation  must  be  absent.  Since  the 
power  is  a  function  of  the  time  of  double  the  frequency  of  the 
pressure  or  current,  it  follows  that  it  cannot  be  represented  by  one 
of  a  family  of  vectors  connected  with  those  two  quantities. 

The  important  quantity  to  consider  is  not,  however,  this 
instantaneous  rate  of  doing  work,  but  the  mean  rate,  and  since 
the  two  quantities  are  assumed  to  be  of  constant  amplitude  and 
period,  it  is  sufficient  to  take  the  mean  over  a  single  period  of  the 
alternation.  Now  the  work  done  while  the  small  angle  dO  is  being 
traversed  will  be  the  product  of  the  instantaneous  rate  of  doing 
work  into  that  angle,  and  as  a  whole  period  corresponds  to  the 
angle  2?r,  the  mean  rate  of  doing  work  over  the  period  will  be 

EC  P27r 

given  by  W  =  -=—       sin  0  sin  (0  -  X) .  dO.    But  this  can  be  written 
ZTT  J  o 

as  the  sum  of  two  integrals  by  using  the  substitution  on  page  13, 
that  is  by  considering  the  current  as  having  the  two  components 
respectively  in  phase  with  and  in  quadrature  with  the  pressure. 
We  thus  get 

EC  r27r 

W  =  -= —  I     (sin2  0 .  cos  X  —  sin  0 .  cos  0 .  sin  X) .  dO, 
ZTT  Jo 

,„     EC  f 27r  f  /I  -  cos  20\      sin  20 

or  W  =  ^ -       ^cosX 


But  the  integrals  of  sin  20  and  cos  20  over  a  complete  period  are 
necessarily  zero,  and  hence  we  have 

w      EC  EC 

W  =  -= —  7T  .  COS  X  =  -TC—  COS  X. 

ZTT  2 

With  sine  pressures  and  currents  we  know  that  the  virtual  value 
is  \/2  times  the  maximum,  and  hence  we  finally  have 

W  =  £<&.caa\. 

It  will  be  noted  that  the  integral  corresponding  to  the  quad- 
rature component  of  the  current  contributes  nothing  to  the  power, 
this  is  therefore  usually  called  the  Wattless  component  of  the 
current,  the  other  component  is  the  Power  component  of  the 
current.  Thus  if  the  current  C  cos  (pt  —  X)  be  flowing  under  the 
pressure  E  smpt  the  wattless  component  has  the  value  <^? 


MEASUREMENT   OF    POWER  29 


while  the  power  component  is  <^,  =  '^cos\.  Thus  if  the  quantity 
OA  in  Fig.  11  represents  the  pressure  in  a  circuit  and  OB  is 
the  current,  the  dotted  curves  will  be  these  two  components  of 
the  current. 

Mean  power  in  a  vector  representation.  Although,  as 
previously  mentioned,  the  instantaneous  power  cannot  be  repre- 
sented by  a  vector,  yet  when  the  two  vectors  of  pressure  and 
current  are  given  it  is  possible  to  represent  the  mean  power  as 
follows.  Consider  Fig.  11  and  let  OA  be  the  vector  representing 
the  pressure,  that  is  one  whose  length  is  the  maximum  value  of 
the  pressure  or  E,  and  let  OB  represent  the  corresponding  current, 
that  is  OB  is  equal  to  the  maximum  current  C  ;  then  the  angle 
A  OB  is  the  phase  angle,  \.  It  has  just  been  seen  that  the  mean 
power  is  given  by  the  expression  JEC  cos  \  and  in  this  case  the 
projection  OC  of  OB  on  0  A  is  the  value  of  C  cos  X,  hence  the  mean 
power  will  be  representable  by  one  half  the  product  of  these  two 
lengths  interpreted  on  an  appropriate  scale.  If  the  vectors  are 
drawn  to  give  the  virtual  values  instead  of  the  maximum  ones, 
the  product  gives  the  mean  power  directly. 

In  the  case  where  various  currents  are  flowing  due  to  the  same 
pressure,  it  follows  that  the  relative  mean  powers  will  be  propor- 
tional to  the  lengths  of  the  projections  of  the  current  vectors  on 
the  pressure  one.  It  is  important  not  to  confuse  the  original 
vectors  with  this  product,  which  is  what  is  called  the  "  scalar 
product  "  of  the  original  vectors  ;  such  a  product  is  a  mere  number 
and  has  no  vectorial  or  directed  properties. 

In  some  cases  it  will  be  found  more  convenient  to  let  the 
projection  of  the  pressure  vector  on  the  current  one  be  taken 
as  a  measure  of  the  power,  but  it  is  evident  that  no  essential 
difference  exists  between  the  two  methods. 

Power  factor.  We  see,  then,  that  with  alternating  currents 
the  product  of  the  pressure  £  and  current  ^  give  no  indication  of 
the  mean  power  that  is  being  produced;  that  product  must  be 
multiplied  by  a  factor,  cos\,  to  determine  the  power.  This 
multiplier  is  known  as  the  Power  Factor  of  the  circuit  in  which 
the  current  flows.  The  product,  £r$,  is  often  called  the  apparent 
power,  and  since  it  does  not  represent  (in  general)  a  true  power, 
it  should  not  be  designated  by  the  term  Watts.  In  cases  where 
the  apparent  power  is  spoken  of,  it  is  said  to  be  reckoned  in  volt-  • 
amperes  or  kilo-  volt-amperes  as  the  case  may  be. 

Equivalent    simple    harmonic    pressure    and    current. 

Since  in  very  many  cases  the  currents  and  pressures  are  non- 
sinusoidal  it  is  of  importance  to  consider  the  assumptions  that 
must  be  made  in  order  that  we  may  treat  them  as  if  they  were 
so,  since  the  sine  case  is  so  easily  dealt  with  by  means  of  analysis 


30 


ALTERNATING   CURRENTS 


or  vectors.  Take  for  example  the  two  curves  in  Fig.  21  as 
representing  the  case  of  a  non-sinusoidal  pressure  and  current. 
As  far  as  concerns  the  representation  of  the  current,  we  could  find 


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Fig.  21. 

the  mean  square  of  its  ordinates  and  construct  a  sine  curve  with 
an  amplitude  equal  to  the  root  of  this  mean  square  multiplied  by 
V2,  and  we  could  take  this  curve  or  its  corresponding  vector  to 
represent  the  quantity  under  consideration.  In  a  similar  manner 
the  actual  pressure  curve  could  be  replaced  by  an  equivalent  sine 
one.  But  this  does  not  tell  us  the  angle  of  phase  relation  at  which 
we  must  draw  the  two  curves ;  this  point  can  be  settled  as  follows. 
The  actual  mean  power  can  be  found  by  drawing  the  curve  of 
instantaneous  power  and  taking  the  mean  ordinate.  Then  if  the 
two  equivalent  sine  curves  be  drawn  at  such  a  phase  angle  that 
.the  mean  power  they  represent  is  the  same  as  this  amount,  we  can 
say  that  the  two  sine  curves  can  be  taken  instead  of  the  actual 
ones.  This  angle  of  phase  difference  being  A,  we  can  call  it  the 
equivalent  angle  of  phase  difference  and  its  cosine  will  be  the 
power  factor.  In  Fig.  22  the  two  curves  of  pressure  and  current 
are  sines  having  the  same  virtual  value  as  those  on  Fig.  21.  The 
power  curve  is  drawn  at  such  a  phase  angle,  A,,  as  to  give  the  same 
mean  power  as  that  of  the  original  curves. 


MEASUREMENT   OF   POWER 


31 


It  should  be  noted  that  the  angle  \  has  no  existence  on  the 
original  curves.  It  is  not  the  angle  between  their  zero  values  nor 
between  their  maximum  ones,  in  fact  these  would  in  general  be 
different;  hence  it  must  be  bonie  in  mind  that  with  non-sinusoidal 
quantities  the  expression  "cos  X"  must  be  taken  as  merely  meaning 
the  power  factor  of  the  circuit,  that  is  the  number  by  which  the 


900-,  30 


^€00 


I" 

20 


30 


Cur 


fit 


Fig.  22. 

product  of  the  virtual  pressure  and  current  must  be  multiplied  in 
order  to  give  the  mean  power,  and  if  we  wish  to  consider  the 
angle  X  as  a  truly  existing  one  it  can  only  be  so  considered  when 
we  imagine  the  quantities  to  be  replaced  by  their  equivalent 
harmonic  representatives.  It  then  comes  to  this,  that  an  alter- 
nating pressure  and  the  corresponding  current  can  be  replaced  by 
two  sine  quantities  provided  these  have  the  same  virtual  values 
and  represent  the  same  mean  power  as  the  actual  pressure  and 
current. 

It  may  be  noted  that  the  only  possible  case  in  which  the 
power  factor  is  unity  is  that  in  which  the  current  and  pressure 
curves  have  the  same  form.  Let  c  and  e  be  the  instantaneous 
values  of  the  same,  the  virtual  values  being  <$  and  £,  so  that  ^2  is 
the  mean  value  of  c2  and  £*  that  of  &.  Consider  the  expression 
(<@e  —  <£c)2 ;  being  a  square  it  is  positive  and  hence  we  have 
<@* .  ez  +  £* .  c2  is  greater  than  2 .  ec .  S^.  This  being  instan- 
taneously true  is  also  true  for  the  means,  and  if  we  denote  the 


32  ALTERNATING   CURRENTS 

mean  value  of  ec  by  W  we  must  then  have  2 .  S- .  ^'2  is  greater 
than  Z.S.yg.W.  Hence  under  all  circumstances,  except  when 
^e  —  £c,  the  mean  power  is  less  than  the  product  & .  *$.  This 

condition  evidently  reduces  to  -  =  ^  or  that  the  ratio  of  the  current 

C  K> 

and  pressure  is  at  every  instant  constant,  that  is  the  curves  have 
the  same  shape. 

It  was  noted  on  p.  15  that  the  sole  condition  that  will  ensure 
that  the  shape  of  the  current  curve  is  identical  with  that  of  the 
pressure  one  is  that  the  circuit  should  contain  an  ohmic  resistance 
and  nothing  else  whatever.  For  in  this  case,  as  we  saw,  the 
current  at  any  instant  is  exactly  equal  to  the  pressure  at  that 
instant  divided  by  the  resistance.  Under  no  other  circumstance 
can  a  circuit  be  without  a  power  factor.  Hence  if  in  any  case  it  is 
necessary  to  provide  a  circuit  in  which  the  current  and  pressure 
curves  are  exact  copies  the  one  of  the  other  this  can  only  be 
secured  by  taking  care  that  the  circuit  has  ohmic  resistance  alone 
and  is  quite  devoid  of  induction  or  capacity. 

The  replacement  of  any  arbitrary  pressure  and  the  corresponding 
current  by  a  pair  of  equivalent  sinusoidal  ones  in  the  case  just 
described  can  always  be  effected  in  the  manner  considered,  but  it 
will  not  follow  that  such  a  replacement  by  a  system  of  plane 
vectors  can  always  be  found.  Consider  the  case  where  a  non- 
sinusoidal  pressure  of  the  instantaneous  value  e  is  acting  on  a 
circuit  of  constant  resistance  R  and  constant  self-induction  L,  the  re- 

dc 

lation  giving  the  current  will  then  be  e  =  cR  +  L  -^  .    Now  let  the 

at 

dc 

instantaneous  pressure  cR  be  given  by  ^  and  the  pressure  L  -%-  by  e2. 

otv 

Deduce  the  virtual  value  of  the  current  and  let  the  equivalent  sine 
be  represented  by  OC  (Fig.  23),  the  equivalent  sine  for  el  will  be  given 


by  OR  in  phase  with  OC,  and  will  be  determined  in  value  by  the 
fact  that  half  the  product  of  the  maximum  of  this  equivalent  sine 


MEASUREMENT   OF   POWER  33 

into  the  maximum  of  the  current  given  by  OC  must  represent  the 
total  power  wasted  in  the  circuit.  Now  the  equivalent  sine  for 
the  E.M.F.  e2  must  evidently  be  found  from  the  fact  that  it  corre- 
sponds to  no  waste  of  energy  and  hence  will  be  given  by  a  vector 
RL  drawn  perpendicular  to  OR,  and  of  such  a  length  as  to  repre- 
sent the  value  of  the  maximum  E.M.F.  of  the  sine  curve  equivalent 
to  the  actual  curve  of  E.M.F.  given  by  e.  These  vectors  will  lie 
in  a  plane,  the  angle  ORL  being  a  right  angle,  and  the  angle  LOO 
will  be  a  definite  angle,  X,,  such  that  its  cosine  is  the  power  factor. 
Now  determine  in  the  same  way  the  value  of  the  maximum  of  the 
sine  equivalent  to  the  E.M.F.  e.  It  must  fulfil  the  conditions  that 
it  has  the  same  virtual  value  as  the  actual  curve  for  e  and  gives 
the  same  power  with  the  current  vector  OC  as  the  actual  pressure 
and  current  give.  Let  the  value  be  OP  and  the  phase  angle  X. 
It  does  not  follow  that  the  length  OP  will  be  equal  to  the  line  OL 
or  that  the  angles  X  and  Xa  are  the  same,  in  fact  this  will  never  be 
exactly  the  case.  If  a  perpendicular  be  drawn  at  L  to  the  plane 
ORL  it  will  be  found  that  OP  will  lie  on  this  perpendicular  so 
that  the  projection  of  OP  on  that  original  plane  is  OL.  Hence 
the  lines  OP  and  OG  correspond  to  the  maximum  of  the  equivalent 
sines  of  pressure  and  current  given  in  Fig.  22.  The  lines  OP,  PR 
and  OR  evidently  lie  in  a  plane  and  PR  is  perpendicular  to  OR  and 
may  hence  be  taken  in  that  plane  to  represent  the  inductive  effect. 

It  will  thus  be  seen  that  the  true  representation  of  all  the 
quantities  in  the  case  where  the  pressures  or  currents  are  non- 
sinusoidal  is  of  necessity  one  which  cannot  be  reproduced  fully  in 
a  plane  figure,  but  must  be  referred  to  three  dimensions. 

In  some  cases,  such  as  those  concerned  with  the  reactive  effects 
of  the  armature  of  a  dynamo,  the  value  of  the  quantity  corre- 
sponding to  the  self-induction  L  is  not  constant,  and  it  will  follow 
that  in  such  cases,  even  if  the  E.M.F.  impressed  on  the  circuit 
is  a  sine  one,  the  current  will  not  be  so,  and  similar  considerations 
will  again  be  brought  into  play.  In  most  practical  cases,  however, 
the  sets  of  vectors  are  sufficiently  nearly  in  a  plane  to  prevent  any 
serious  error  being  made  by  neglecting  the  difference  between  the 
angles  X  and  Xj. 

Theorem  on  mean  values.  The  following  relation  is  of 
importance  for  some  considerations  and  refers  to  any  two  periodic 
quantities  whatever  the  form.  Let  one  of  these  have  the  in- 
stantaneous value  a  and  the  other  the  instantaneous  value  6. 
We  must  have  the  relation 

d      ,  db      ,    da 

dfab  =  a-di  +  b'Tf 

Hence  if  the  two  sides  be  integrated  over  a  period  it  follows  that 

1  [T      db  ,       1  [r ,    da      1  [T  d      ,     ,,     1  fr  , 

-      a.-rrdt+-\   b.-j-  =  -l   -j-.ab.dt  =  -     d(ab\ 

TJo      dt          TJo       dt      rJodt  rJo 

L.  3 


34  ALTERNATING   CURRENTS 

but  the  right-hand  side  is  of  necessity  zero  since  the  values  of  a 
and  b  (and  hence  that  of  the  product  ab)  are  from  the  nature  of 
those  quantities  the  same  at  the  time  zero  and  at  the  time  T, 
we  therefore  have 


1  [T      db  If*      da 

-      a.-j-.dt  =  —      b.-rr- 
rJo       dt  rJo       dt 


It  follows  that  for  a  single  periodic  quantity,  x,  we  must  always 
have 


'  dt 


MEASUREMENT  OF   ALTERNATE   CURRENT   POWER. 

The  wattmeter.  When  considering  the  dynamometer  we 
saw  that,  from  the  construction  of  the  instrument,  the  angle  of 
torsion  of  the  spring  was  related  to  the  product  of  the  currents  by 
the  equation  mean  (c^)  =  k .  a,  where  d  and  c2  are  the  instanta- 
neous currents  in  the  two  coils.  Let  such  an  instrument  be 
connected  as  shown  in  Fig.  24,  where  X  is  the  load  in  which  the 


O 


Fig.  24. 

power  is  to  be  measured  and  R  is  a  high  non-inductive  resistance 
in  series  with  one  of  the  coils  (usually  the  movable  one)  which  is 
placed  as  a  shunt  on  the  load,  the  other  coil  being  in  series  with 
it.  Let  e  denote  the  pressure  at  the  terminals  of  the  load,  cm 
the  current  in  the  main,  c8  that  in  the  shunt.  From  the  law  of 
the  instrument  we  have  mean  (cmcs  +  cs2)  =  ka  since  the  current 
in  the  shunt  coil  is  cg  while  that  in  the  series  coil  is  cm  +  cs. 
It  follows  that  on  multiplying  each  side  by  R  we  have 

mean  (cmcsR  +  cs*R)  =  kRa. 

Now  mean  (cs2R)  is  the  loss  of  power  in  the  shunt  circuit ;  further, 
since  the  shunt  circuit  has  been  arranged  so  as  to  be  non-inductive, 
at  every  instant  we  have  e  =  csR.  Thus  if  W  is  the  power  given 
to  the  load  and  W8  that  given  to  the  shunt  we  have 


MEASUREMENT  OF  POWER  35 

But  if  the  resistance  of  the  shunt  be  high  we  can  arrange 
matters  so  that  the  loss  in  it  is  small  compared  to  the  power  that 
has  to  be  measured.  In  any  case  a  correction  could  be  applied  to 
an  observed  reading  if  the  pressure  employed  and  the  resistance 
of  the  shunt  be  known.  The  error  due  to  the  shunt  current 
flowing  round  the  series  coil  can  be  annulled  as  follows.  Let  an 
additional  coil  be  wound  over  the  series  coil  possessing  exactly  the 
same  number  of  turns  as  that  coil  but  placed  in  series  with  the 
shunt  circuit,  and  with  the  shunt  current  passing  round  it  the 
opposite  way  to  the  main  current  in  the  series  coil.  It  is  evident 
that  the  ^magnetic  effect  of  such  a  coil  will  just  annul  that 
produced  by  the  shunt  current  passing  in  the  series  coil  and 
thus  the  instrument  will  read  the  power  correctly.  Such  a  coil 
is  called  a  compensator. 

Another  method  of  connection  is  as  shown  in  Fig.  25.     In  this 
case   we   evidently   have    mean  (cmc8.  R)  =  k.  R.a.     But   if    the 


Fig.  25. 

resistance  of  the  series  coil  be  S,  we  see  that  e  +  cmS  =  csR. 
Hence  mean  (cme  +  cm2$)  —  k.R.oL.  Or  if  W  be  the  power  given 
to  the  circuit  and  Wm  that  lost  in  the  series  coil,  we  have 

W+  Wrn  =  k.R.oL. 

Thus  with  a  direct  connection  to  the  circuit  we  will  measure  too 
large  a  power  by  the  loss  in  either  of  the  coils  of  the  instrument ; 
which  method  is  the  better  will  depend  on  the  nature  of  the 
circuit.  If  we  leave  out  this  small  error  we  can  write  W=k.R.a, 
or  if  the  resistance  of  the  shunt  be  constant  under  all  conditions, 
W=K . a..  Such  an  instrument  is  known  as  a  Wattmeter.  From 
the  way  in  which  we  have  deduced  its  law  it  is  evident  that  the 
calibration  and  constant  K  are  the  same  both  for  direct  and  for 
alternating  power  even  when  the  latter  is  of  any  wave  form.  For 
the  purpose  of  determining  the  constant  it  is  not  necessary  to 
actually  waste  the  power  corresponding  to  any  required  reading, 
the  main  current  can  be  supplied  from  one  source  and  measured 
by  an  ammeter  while  the  pressure  is  applied  to  the  shunt  circuit 
from  another  and  measured  by  a  voltmeter. 

Wattmeter  error.     Throughout  the  above  we  have  assumed 
that  the  current  in  the  shunt  circuit  is  an  exact  copy  of  the 

3—2 


36 


ALTERNATING   CURRENTS 


pressure  at  its  terminals,  and  it  is  only  under  such  circumstances 
that  the  wattmeter  will  measure  the  true  value  of  the  mean 
power.  If  the  pressure  and  current  be  sine  quantities  we  can 
investigate  the  effect  of  the  presence  of  a  phase  difference  between 
the  current  in  the  shunt  and  the  pressure  at  its  terminals  as 
follows.  Let  OA,  Fig.  26,  be  a  vector  representing  the  pressure  at 


Fig.  26. 

the  terminals  of  the  load.  Let  OAB  be  the  impedance  triangle  for 
the  shunt  circuit  in  which  a  small  self-induction  is  supposed  to  be 
present,  and  let  OAG  be  the  same  triangle  for  the  load  which  is 
also  taken  as  inductive.  Let  R  be  the  resistance  of  the  shunt  and 
r  that  of  the  load.  Then  the  maximum  currents  in  the  two 

circuits  will  be  -^  and  -  -  respectively.     Hence  the  mean  couple 

./L  / 

that  the  instrument  measures  will  be  evidently  proportional  to 

-„  .~OC.~OB .  cos (\  -  <f>). 
rR 

But  we  wish  to  measure  the  mean  power,  that  is,  the  mean 
product  of  the  currents  -=r-  and  ^  .  Hence  the  true  reading  of 
the  wattmeter  should  be  proportional  to 

-Vo5.oa.cosx. 

rH 


MEASUREMENT   OF   POWER  37 

Hence  in  this  case  we  must  multiply  the  observed  reading  of  the 
wattmeter  by  the  factor 

Qq.aa.  cos  x 


in  order  to  get  what  its  reading  would  have  been  if  the  shunt  had 
been  non-inductive.  But  we  see  that  OB  =  OA  .  cos  <£.  Hence 
the  correction  factor  becomes 

cos  X  cos  X 

or 


cos  </> .  cos  (X  —  (/>)        cos-  <f> .  cos  X  +  sin  <f> .  cos  <f> .  cos  X 

sec2  6  1  +  tan2  <f> 

or  ^ r  r  or 


1  +  tan  <f>  tan  X        1  +  tan  </>  .  tan  X  * 

This  can  be  written  in  terms  of  the  resistances  and  self-inductions 
as  follows, 


where  Z1,  Rl  refer  to  the  shunt  circuit  and  L  and  R  to  the  load. 

It  will  be  seen  that  the  only  way  to  make  the  correction 
independent  of  the  periodicity  and  of  the  nature  of  the  load  is  to 
cause  jLj  to  be  zero.  This  is  the  case  we  have  already  taken  and 
hence  we  see  that  it  is  necessary  to  have  the  shunt  circuit  possessed 
only  of  ohmic  resistance.  This  is  a  condition  which  cannot  be 
carried  out  with  absolute  accuracy.  The  ordinary  double  wound 
coil  as  a  matter  of  fact  usually  has  a  considerable  capacity  effect 
owing  to  the  method  of  winding  being  such  that  the  ends  have  a 
high  pressure  between  them.  A  close  approximation  to  the 
desired  result  can  be  attained  by  winding  the  shunt  resistances  on 
thin  but  stiff  sheets  of  wood  or  millboard  and  making  them  of 
wire  of  very  high  specific  resistance.  Another  method  is  to  use  a 
sort  of  ribbon  for  the  shunt  resistance  in  which  the  web  is  an 
insulating  thread  while  the  woof  is  a  fine  high  resistance  wire. 
Another  point  that  must  be  borne  in  mind  is  that  we  have 
assumed  that  no  magnetic  fields  other  than  those  produced  by  the 
shunt  and  series  coils  are  present.  Such  other  fields  would  be 
produced  if  any  coils  carrying  the  current  were  near  the  instru- 
ment, and  also  by  the  presence  of  fields  due  to  the  induction  of 
alternating  currents  in  any  metal  near  the  instrument  by  the 
currents  circulating  in  the  coils  of  the  instrument  itself.  It  follows 
that  all  metallic  parts  (other  than  the  coils  themselves  and  the 
suspensions)  should,  in  cases  where  considerable  accuracy  is  re- 
quired, be  carefully  avoided. 

The  wattmeter  we  have  so  far  considered  requires  that  the 
coil  should  be  brought  back  to  its  zero  position  at  each  reading. 
In  fact  it  is  a  zero  reading  instrument.  In  very  many  cases  this 
is  not  convenient.  If  the  fine  wire  coil  be  suspended  about  an 


38 


ALTERNATING    CURRENTS 


axis  and  provided  with  proper  controlling  couple  it  is  evident  that 
the  angle  through  which  the  coil  deflects  could  be  taken  as 
measuring  the  power.  As  a  rule  such  a  wattmeter  is  not  quite  so 
accurate  as  the  zero  reading  one,  but  the  convenience  of  direct 
reading  more  than  compensates  for  this  in  practical  cases.  A 
wattmeter  depending  on  electromagnetic  action  which  is  unaffected 
by  external  fields,  and  which  has  the  advantage  of  producing  much 
larger  deflecting  forces  than  the  type  we  are  considering,  will  be 
referred  to  on  p.  81. 

Three  voltmeter  and  ammeter  methods.  In  the  absence 
of  a  wattmeter  the  following  methods  of  measuring  alternate 
current  power  are  available.  Let  the  circuit  in  which  the  power 
is  to  be  measured  be  placed  in  series  with  a  known  non-inductive 
resistance  and  let  three  voltmeters  be  placed  across  the  whole  and 
each  part  of  the  circuit  formed,  as  shown  in  Fig.  27  at  V1}  F2,  F3. 


Fig.  27. 

Let  R  be  the  value  of  the  non-inductive  resistance,  and  let  el  be 
the  pressure  at  any  moment  between  the  ends  of  the  inductive 
circuit,  e2  that  on  the  non-inductive,  and  es  across  the  whole. 
Then  at  any  instant  we  evidently  have  el  +  e2  —  eS)  and  hence 
#102  =  i  W  —  e\  —  022)-  But  the  instantaneous  power  is  given  by 
the  product  of  the  pressure  el  into  the  current,  that  is  by 


w  = 


j  ^T. 
and  thus  w 


2R 


?  -  e^  -  e*). 


$2,  and  £3  denote  the  readings  of  the  three  voltmeters,  that 
is  the  root  of  the  mean  square  pressures,  and  if  W  denote  the  mean 
power  that  is  being  delivered  to  the  circuit,  it  will  be  readily  seen 
that 

W  =  ^R  ^  ~  ^  ~  ^ 

In  many  cases  it  is  difficult  to  obtain  a  known  non-inductive 
resistance  for  R.  Instead  of  this  we  can  use  some  ordinary  incan- 
descent lamps  but  in  this  case  the  value  of  R  will  be  unknown ; 
in  order  to  find  that  value  an  ammeter  can  be  placed  in  series 


MEASUREMENT   OF   POWER 


39 


with  them,  and  from  the  reading  of  this  and  the  voltmeter  we  can 
find  the  value  of  R. 

It  can  readily  be  seen  that  the  value  of  W  is  most  accurately 
determined  when  Si  and  £z  are  equal,  and  hence  in  such  a  case 
this  method  necessitates  the  use  of  a  pressure  considerably  in 
excess  of  that  required  for  the  operation  of  the  apparatus  under 
test ;  in  many  cases  this  is  difficult  to  procure.  The  following 
form  of  the  experiment,  in  which  the  non-inductive  resistance  and 
the  circuit  under  test  are  put  in  parallel  (see  Fig.  28)  and  the 


Fig.  28. 

currents  taken  by  the  two  combined  and  each  separately  are 
measured,  only  necessitates  the  flow  of  a  larger  current,  and 
permits  the  use  of  the  normal  pressure  for  the  test. 

Let  the  virtual  currents  as  shown  by  the  ammeters  Al}  A2,  and 
A3  be  respectively  ^,  ^2,  and  ^3,  then  it  will  readily  be  seen  that 

7} 

the  mean  power  is  given  by  W  =  -  (^32  -  <@*  -  <^22),  where  R  is 

as  before  the  value  of  the  non-inductive  resistance.  If  this  consists 
of  incandescent  lamps  the  value  of  R  can  be  found  by  placing 
a  voltmeter,  preferable  a  hot  wire  one,  across  the  non-inductive 
resistance,  and  neglecting  the  small  drop  in  the  ammeter  A2. 

Modified  3  voltmeter  method.  The  following  modification 
of  the  three  voltmeter  method  enables  the  measurement  of  small 
phase  angle  to  be  made.  Let  it  be  desired  to  measure  the  phase 
angle  between  the  current  and  pressure  in  the  coil  EC  (Fig.  29). 
Connect  in  series  with  it  a  non-inductive  resistance  AB  and  in 
parallel  with  both  a  resistance  RR  along  which  a  sliding  contact 
can  move.  The  vector  figure  of  the  pressures  will  then  be  as 
given  below,  being  such  that  AB  is  the  pressure  on  the  series 
resistance,  BC  that  on  the  terminal  of  the  coil,  and  the  third  AC 
that  between  the  terminals  of  R.  Let  Q  be  any  point  on  this 
resistance,  then  the  pressure  between  B  and  Q  will  be  given  by 
the  line  BQ.  Of  all  possible  points  on  R  there  will  be  one, 
namely  P,  which  is  such  that  BP  is  perpendicular  to  AC  and 
the  position  of  that  point  will  be  shown  by  a  voltmeter  joining 


40 


ALTERNATING   CURRENTS 


B  to  the  sliding  contact  on  R  giving  a  minimum  reading.    Let  £m 
be  that  minimum  reading,  £  the  pressure  on  the  coil,  £^  that  on 


Fig.  29. 

the  resistance  AB,  then  if  the  angles  be  as  shown,  X  being  the 
desired  phase  angle,  we  evidently  have 


Hence  the  phase  angle  can  readily  be  found. 

This  method  has  the  advantage  that  it  is  not  necessary  to  have 
any  large  drop  along  the  series  resistance,  and  hence  the  ordinary 
supply  pressures  are  in  general  sufficient.  In  such  a  case  when 
the  drop  £m  is  but  a  small  fraction  of  the  pressure  £  it  is  evident 
that  /3  is  very  small  compared  with  the  other  angles,  and  hence 
the  two  observations  of  £m  and  £^  will  give  X  with  fair  accuracy. 

This  method  depends  on  the  possibility  of  providing  suitable 
low  reading  alternate  current  voltmeters,  and  in  cases  where  £m 
is  very  small  indeed,  instruments  of  the  proper  type  are  not  in 
general  available.  The  method  of  procedure  in  this  case  will 
be  found  on  p.  221. 

Electrometer  methods.  The  properties  of  the  quadrant 
electrometer  can  be  used  to  give  a  method  of  measuring  power, 
and  with  proper  precautions  this  method  is  a  very  good  one  to 
employ,  the  difficulty  encountered  in  getting  the  shunt  circuit 
of  a  wattmeter  entirely  devoid  of  lag  or  lead  is  avoided,  and  thus 
measurements  with  very  low  power  factors  can  be  readily  carried 
out.  Let-  Fig.  30  represent  a  quadrant  electrometer  in  which  the 
pressures  between  the  needle  and  the  two  pairs  of  quadrants  are 


MEASUREMENT   OF   POWER 


41 


as  shown,  being  Vl  between  one  pair  and  the  needle  and  V9 
between  the  other  pair  and  the  needle.  If  the  instrument  be 
properly  designed,  both  as  regards  the  form  of  the  quadrants  and 
that  of  the  needle,  and  also  as  regards  the  control,  we  can  show 
that  the  relation  between  these  pressures  and  the  angle  a,  through 
which  the  needle  turns,  is  given  by 

(y?-Vf)  =  k*. 

The  needle  N  forms  a  condenser  of  variable  capacity  with 
each  of  the  cross  connected  pairs  of  quadrants.  If  the  instrument 
be  made  in  a  perfectly  symmetrical  manner  it  is  evident  that 
when  the  needle  moves  through  the  angle  a  from  its  position 
of  rest,  the  portions  of  its  area  that  emerge  from  within  one  pair 
of  quadrants  and  enter  into  the  other  pair  are  exactly  equal  and 


Fig.  30. 

that  each  portion  is  proportional  to  that  angle  of  twist.  Hence 
the  capacity  of  the  condenser  formed  by  the  needle  and  one  pair 
of  quadrants  will  increase  by  a  definite  amount,  while  that  formed 
by  the  other  pair  will  decrease  by  the  same  amount,  and  further 
this  amount  can  be  written  as  /.a  where  /is  a  constant,  being  in 
fact  the  capacity  per  radian  of  the  condenser  formed  by  either 
part  of  quadrants  and  the  needle.  Hence  if  pressures  Fx  and  F2 
be  applied  as  shown,  from  what  was  said  on  p.  8  the  change  of 
energy  due  to  the  resulting  angular  twist  a,  will  be 


If  the   suspension  be   such  as  to  give  a  controlling   couple 
proportional  to  the  angle  of  twist,  as  is  the  case  for  a  torsional 


42 


ALTERNATING   CURRENTS 


suspension,  and  nearly  so  for  a  very  long  bifilar,  the  work  done 


due  to  twisting  the  suspension  will  evidently  be  given  by  —  -— 

2^ 

where  a  is  a  constant.    Thus  on  equating  these  amounts  of  energy 
change  and  work  done,  we  get 


It  will  be  found  that  most  electrometers  do  not  fulfil  this 
relation  with  sufficient  accuracy  to  enable  it  to  be  assumed  as 
a  basis  for  developing  a  method  of  power  measurement.  This 
point  may  be  tested  by  connecting  the  needle  to  one  pair  of 
quadrants,  thus  reducing  one  of  the  potential  differences  to  zero, 
and  putting  known  pressures  across  the  needle  and  the  free  pair. 
Since  the  expression  then  becomes  Fi2  =  ka  it  should  be  found 
that  the  deflection  is  rigidly  proportional  to  the  square  of  this 
pressure  over  the  whole  desired  range  of  the  instrument.  If  this. 
is  not  the  case,  the  instrument  is  not  suited  for  the  purpose 
we  are  going  to  consider.  The  principal  difficulty  arises  from 
want  of  symmetry  which  not  only  prevents  the  capacity  altering 
exactly  proportional  to  the  angle,  but  also  introduces  forces  of 
attraction  between  the  needle  and  the  quadrants,  or  other  parts 
of  the  apparatus,  which  we  have  not  considered,  and  which 
virtually  prevent  the  controlling  force  being  due  solely  to  the 
suspension  as  we  have  taken  it  to  be.  To  enable  these  outstanding 
quantities  to  be  reduced  to  a  minimum  it  has  been  found  best 
not  to  aim  at  a  very  sensitive  instrument,  as  is  ordinarily  done, 
but  to  have  one  with  a  fairly  heavy  needle  so  that  any  slight 
want  of  symmetry  will  produce  a  relatively  small  effect,  and  also 
to  make  the  parts  with  larger  distances  apart  than  is  ordinarily 
employed.  When  all  precautions  are  taken  it  is  possible  to  obtain 
an  instrument  that  obeys  the  desired  law  within  a  small  fraction 
of  one  per  cent. 

Let  such  an  electrometer  have  the  value  of  its  constant  k 
determined  as  just  described,  and  let  it  be  desired  to  use  it  for  the 


Fig.  31. 

purpose  of  power  measurements.    Let  the  load  AB,  Fig.  31,  be  put 
in  series  with  a  known  resistance  BC  of  value  R,  which  for  alternate 


MEASUREMENT   OF   POWER 


43 


currents  must  be  quite  non-inductive.  Consider  first  that  direct 
currents  are  being  used,  and  join  up  as  shown  with  the  quadrants 
connected  to  the  ends  of  the  standard  resistance  and  the  needle  to 
the  other  end  of  the  load.  Let  V  be  the  pressure  across  the  load, 
and  v  that  across  the  resistance,  then  the  pressure  between  the 
needle  and  one  pair  of  quadrants  is  V  while  that  between  the 
needle  and  the  other  is  V  +  v.  Thus  if  the  quadrant  electrometer 
deflects  through  the  angle  a,  the  connection  between  the  different 
quantities  will  be 

(V+vY-V*  =  k.oL. 

This  reduces  to  2  Vv  +  v2  =  k  .  a. 

But  since  v  is  due  to  the  current  G  flowing  through  R  we  have 

v  =  C.R, 
and  hence  we  have  finally 


But  VC  is  the  power  that  is  being  supplied  to  the  circuit  while 
%C2.  R  is  that  lost  in  the  little  series  resistance,  and  if  the  latter 
is  small,  the  deflection  will  be  nearly  proportional  to  the  power. 
If  the  source  of  power  be  alternating  and  the  small  resistance  be 
non-inductive,  it  is  evident  that  the  same  is  true  for  the  mean 
power,  and  hence  the  electrometer  will  act  as  a  wattmeter. 

The  following  methods  involve  the  use  of  a  transformer  and 
will  be  better  understood  later  on.  Take  the  case  where  the 
supply  is  at  low  pressure.  Let  the  current  be  passed  through  the 
primary  of  a  transformer  T,  the  secondary  of  which  is  closed  on  a 
resistance  of  amount  R,  and  let  the  connections  be  made  as  in 
Fig.  32.  We  must  at  present  assume  that  the  transformer  operates 


Fig.  32. 

in  such  a  way  that  if  c  be  the  current  in  the  main  circuit,  that  in 
the  secondary  will  be  a .  c  where  a  is  a  constant.    Hence  since  R  is 


44 


ALTERNATING   CURRENTS 


non-inductive  the  instantaneous  pressure  at  its  terminals  will  be 
e  =  a.R.c.  Let  the  corresponding  instantaneous  pressure  on  the 
load  be  e.  Then  the  pressure  between  one  pair  of  quadrants  and 

the  needle  will  be  e  +  •=  while  that  between  the  other  pair  of  quad- 
rants and  the  needle  will  be  e  —  -= .  The  virtual  value  of  the  two 

2 

pressures  e  and  e  would  be  shown  by  voltmeters  at  V  and  v.  In 
this  case  the  instantaneous  couple  will  be  given  by 

Ce\2       /         €\2 
e  +  ~  J  —  (e  —  n)  =  k  .a.  or  e  .e  =  k  .a. 

But  owing  to  the  inertia  of  the  needle  the  mean  deflection  will 
indicate  the  mean  value  of  the  couple,  and  hence  we  have  the 
mean  power  as  given  by 

W  =  mean  ce  =  — ^  . «. 
an 

In  the  case  where  the  supply  is  at  very  high  pressure  the 
connections  shown  in  Fig.  33  can  be  used,  where  the  power  is 


Fig.  33. 

transmitted  to  the  load  QS  by  means  of  a  transformer  in  a  manner 
yet  to  be  considered.  The  needle  is  attached  to  a  point  P 
on  the  secondary  of  this  transformer  so  that  the  pressure 
between  0  and  P  is  only  I /nth  of  that  between  0  and  T.  RI  is 
a  non-inductive  resistance  of  the  value  Rl}  and  R2  is  a  second 

whose    resistance   is   Rl  (•=  —  1 


As   before   let   e   be   the  in- 


stantaneous drop  down  Rl}  then  we  have  e  =  cRl.     Further  the 
pressure  between  Q  and  S  being  e,  that  between  0  and  S  is  e  +  e 


while  that  between  0  and  T  is  e  + 


n.e 


Hence  the  pressure 


between  0  and  P  is  -  -I-  ^  while  between  Q  and  P  it  is -=  • 

Thus  the  instantaneous  relation  is  in  this  case  —  —  k.a  and  thus, 


MEASUREMENT   OF    POWER 


45 


nk 


as  in  the  last  one,  the  mean  power  is  given  by  W  =  -p- .  a.     It  will 

xil 

be  seen  that  when  n  =  2  the  resistance  R^  can  be  suppressed. 

It  is  sometimes  desired  to  measure  a  "  fictitious  "  power,  that 
is,  to  supply  current  and  pressure  to  the  parts  of  an  instrument 
under  test  from  two  sources  without  wasting  the  power  corre- 
sponding to  their  product.  Thus  suppose  we  wish  to  test  a  supply 
meter  under  different  conditions  of  load  and  phase  angle.  Let  the 
current  be  supplied  by  one  dynamo  and  the  pressure  by  another, 
the  two  being  so  arranged  that  the  armatures  can  be  given  any 
desired  angular  relation,  but  are  rigidly  driven  as  a  whole.  Let 
the  electrometer  be  joined  up  as  shown  in  Fig.  34  with  a  non- 


NWVWWWS/WWVWWWWM 

< V 


Meter 

Fig.  34. 

inductive  resistance  in  series  with  the  current  circuit  and  the 
pressure  circuit  attached  at  one  end  to  the  middle  of  this 
resistance,  and  at  the  other  to  the  needle.  If  R  be  the  value 
of  this  resistance  the  drop  for  any  current  c  will  be  e  =  cR. 
From  the  method  in  which  the  circuit  is  connected  it  will  be  seen 
that  the  pressure  between  the  pairs  of  quadrants  and  the  needle 

are  respectively  e  +  -=  and  e  —  -=  where  e  is  the  pressure  on  the 

shunt  circuit  of  the  meter.  It  follows  that  the  mean  "  power  "  is 
proportional  to  the  deflection  of  the  electrometer.  In  this  way 
many  meters  can  be  tested  at  once ;  it  is  only  necessary  to  put  all 
the  shunts  in  parallel  on  the  one  armature  and  all  the  series  coils 
in  series  with  one  another  on  the  other.  This  method  is  particu- 
larly useful  when  it  is  desired  to  find  the  effect  of  phase  difference 
on  the  action  of  a  meter,  and  to  adjust  the  constants  of  a  large 
number  of  similar  meters  at  the  same  time. 


CHAPTER   IV. 

THE  CHOKING  COIL  AND   TRANSFORMER. 

Choking  Coil.  Ideal  case.  A  choking  coil  consists  of  an 
iron  core  suitably  surrounded  by  a  winding  in  which  the  alternating 
current  flows.  Let  the  dimensions  of  the  apparatus  (see  Fig.  35) 


Fig.  35. 

be  such  that  the  mean  length  of  the  iron  circuit  is  I  centimetres, 
its  cross  section  s  square  centimetres,  and  let  the  relation  between 
the  magnetising  force,  H,  and  the  resulting  induction,  B,  for  the 
iron  of  which  it  is  made  be  as  given  in  Fig.  36.  Let  us  assume 
that  a  current  is  passing  of  such  an  amount  as  to  produce  some 
definite  maximum  value  of  this  induction,  which  we  will  call  B. 
Then  from  the  curve  in  Fig.  36  we  can  find  the  corresponding 
value  of  the  magnetising  force,  let  it  be  H.  If  the  current  be 
alternating,  it  will  produce  an  alternating  flux,  and  as  a  first 
approximation  let  us  take  the  relation  between  the  B  and  the  H 
during  the  cycle  as  being  a  constant  quantity,  namely  BjH, 


THE   CHOKING   COIL   AND   TRANSFORMER 


47 


instead  of  following  a  hysteresis  cycle  as  is  actually  the  case, 
and  further,  let  the  resistance  of  the  winding  be  considered  as 


efuvu 
7.000 
6.000 
5,000 
4-.  000 
3.000 
V.OOO 
1.  OOO 

4 

x 

x 

x 

x 

S 

x 

X 

^ 

/ 

/ 

s 

/ 

/ 

/ 

, 

? 

/ 

/ 

/ 

' 

x 

^^^ 

^ 

~* 

1                            2                           3                          +                          S 

Va.lue     of  H  . 

Fig.  36. 

negligibly  small.     Then  if  C  be  the  maximum  current,  and  T 
the  total  turns  in  the  coil,  we  evidently  have 

(1), 


the  current  being  measured  in  absolute  units.     Again  the  total- 
flux  will  have  a  definite  maximum  value  which  will  be  given  by 
4>  =  Bs.     But  we  can  also  write  pH  =  B,  where  /JL  is  the  value  of 
the  permeability,  assumed  constant.     This  leads  to 


which  could  at  once  have  been  derived  from  the  consideration  of 
the  given  magnetic  circuit.  For  the  reluctance  of  the  circuit  is 

— ,  and  the  magnetomotive  force  is  4?rCT  hence  the  ^.ux  is  given 

by  the  relation  Flux  =  magnetomotive  force  H-  Reluctance  which 
leads  to  the  above  equation.  The  total  f(ux  can  be  found  either 
from  this  latter  equation  or  from  no  (1).  In  general  the  latter  is 
the  more  convenient  method.  In  some  cases  the  magnetising  force 
is  considered  in  terms  of  the  inch  as  unit  of  length.  Thus  we  have 

H  =  — ^  ampere  turns  per  centimetre 


47T 


or  2*02  ampere  turns  per  inch. 


48 


ALTERNATING   CURRENTS 


In  this  case  the  induction  would  be  in  lines  per  square  inch,  that 
is  6*25  times  the  corresponding  absolute  fluxes. 

It  follows  that  with  the  above  assumption,  if  we  have  a  current 
flowing  which  is  given  by  c  —  C  smpt  it  will  be  accompanied  by 
a  core  flux  given  by  (/>  =  O  sin  pt.  Since  this  flux  is  passing  round 
the  iron  core  it  will  pass  through  each  of  the  T  turns  of  the  coil 
wound  thereon,  and  hence  an  E.M.F.  will  be  generated  in  that  coil 

given  by  e  =  —  T-^,  or  e  =  —  p  .  4>  .  T  .  cos  pt.     Hence  the  maxi- 

mum value  of  the  induced  E.M.F.  will  be   E=p.<&.T,  and  the 
corresponding  virtual  value  will  be 


But  this  is  the  only  pressure  existing  in  the  coil  since  the  resistance 
is  zero,  and  hence  the  applied  pressure  must  be  exactly  equal  and 
opposite  to  this.  The  vector  representation  will  be  as  in  Fig.  37, 


& 


M 


Fig.  37. 

OF  is  the  maximum  flux  and  OM  the  corresponding  magneto- 
motive force  while  OG  is  the  -current.  In  the  present  case  these 
vectors  all  point  in  the  same  direction.  OE  is  the  induced  E.  M.  F. 
lagging  a  quarter  period  after  the  flux  while  0  V  is  the  impressed 
pressure,  exactly  equal  and  opposite  to  the  last. 

Core  loss;  angle  of  hysteretic  lead.  We  will  now  see 
what  alteration  is  produced  if  we  take  into  consideration  the  fact 
that  the  true  relation  between  the  flux  and  the  current  is  a  cyclic 
curve.  In  Fig.  38  is  drawn  a  cyclic  curve  of  flux  and  current  for 
a  definite  core ;  and  in  the  adjoining  figure  is  drawn  a  sine  curve 
of  applied  pressure.  Since  the  maximum  total  flux  is  related  to 
the  maximum  induced  pressure  by  the  equation  E  =  p .  <& .  T,  and 
since  we  have  seen  that  the  flux  leads  the  induced  pressure  and 
lags  on  the  applied  pressure  in  each  case  by  90°,  it  will  be  evident 


THE   CHOKING   COIL   AND   TRANSFORMER  49 

that  the  curve  of  flux  is  a  sine  of  amplitude  — ^  times  the  pressure 

one  interpreted  on  the  proper  scale  of  flux  and  lagging  90°  on 
the  pressure  curve,  and  will  be  as  shown  in  the  figure.     If  the 


Flux. 


Fig.  38. 

cyclic  curve  is  drawn  with  the  same  scale  of  ordinates  for  the  flux 
and  with  the  same  axis  as  the  sine  curve  of  flux,  it  is  only  necessary 
to  project  from  the  flux  sine-curve  to  the  cyclic  curve  in  order  to 
determine  the  current  flowing  at  each  value  of  the  flux.  We  can 
then  erect  at  each  point  along  the  horizontal  axis  an  ordinate 
which  will  represent  to  the  proper  scale  the  current  that  is  flowing 
at  that  instant.  It  is  shown  in  the  figure.  It  will  be  noticed 
that  the  effect  of  hysteresis  is  to  cause  the  current  wave  to  be 
non-sinusoidal  in  shape  and  distorted.  Further,  it  is  no  longer  in 
quadrature  with  the  pressure  curve,  and  hence  power  has  to  be 
supplied  from  the  source  of  energy,  as  must  be  the  case  from  the 
existence  of  hysteresis.  We  can  replace  the  actual  current  curve 
by  its  equivalent  sine  curve  in  the  manner  described  on  p.  29. 
This  is  shown  dotted  in  the  figure,  and  it  will  be  seen  that  the 
flux  leads  the  current  by  a  definite  angle  ^,  which  is  called  the 
Angle  of  Hysteretic  Lead.  Further,  its  complement  is  a  definite 
angle  of  lag,  X,  less  than  90°,  between  the  pressure  and  the  current; 
it  follows  that  the  current  has  a  power  component  of  the  amount 
^  =  ^  cos  X,  and  a  wattless  component  of  the  amount  ^  =  ^  cos  X. 
The  amount  of  the  former  can  be  found  for  a  given  iron  core  in 
the  following  manner.  Let  the  relation  between  the  maximum 
induction  B,  and  the  loss  per  cubic  centimetre  per  cycle  in  ergs,  h, 
be  given,  as  shown  in  the  curve  in  Fig.  39  which  refers  to  the 
same  iron  as  that  for  which  the  B — H  curve  was  given  in  Fig.  36. 
Then  if  n  be  the  periods  per  second  and  v  the  volume  of  the 
core  in  cubic  centimetres,  the  loss  will  be  W  =  h .  n .  v .  10~7. 


50 


ALTERNATING   CURRENTS 


Let  the  pressure  at  which  the  choking  coil  is  to  be  supplied  be 
&,  then  the  in-phase  current  necessary  for  this  core  loss  will 
evidently  be  Wj£.  In  the  case  figured  the  total  current  has  a 


** 

<£>  3 

r*+ 

fl 
•-» 

-S? 

& 

i- 

cj 
G 

fc 

Q. 

8,, 

0) 

c 

•  -J 

<0 

«0 

3 

i 

/ 

/ 

/ 

/ 

f 

/ 

/ 

/ 

/ 

/ 

7 

/ 

/ 

/ 

/ 

/ 

/ 

/ 

/ 

<* 

x 

^ 

^ 

1234-S678 

B  in  1000's 

Fig.  39. 

maximum  which  differs  from  that  of  the  equivalent  sine  current. 
It  is  found  that  with  lower  values  of  the  maximum  induction  than 
that  used  in  this  case,  and  in  fact  for  nearly  all  inductions  usualin 
alternate  current  apparatus,  the  maximum  of  the  equivalent  sine 
current  and  that  of  the  actual  current  are  practically  the  same. 
It  follows  that  since  we  can  calculate  in  the  manner  considered  on 
p.  47,  the  maximum  value,  and  hence  the  virtual  value,  of  the 
current  involved  in  the  production  of  a  given  total  flux,  this  same 
maximum  can  be  taken  as  that  of  the  equivalent  sine  current. 
Hence  the  virtual  value  of  the  total  current  is  known.  But  in  the 
manner  just  described  we  can  at  once  determine  the  power 
component  of  that  current,  and  hence  it  is  known  not  only  in 
magnitude,  but  in  phase  relation.  Thus  ^  be  the  virtual 
current  and  ^  its  power  component,  we  evidently  have 

% 

COS  \  —  -F^. 


THE  CHOKING   COIL  AND  TRANSFORMER 


51 


The  angle  of  hysteretic  lead.  It  can  readily  be  seen  that 
the  angle  of  hysteretic  lead  is  fixed  for  a  circuit  when  the  quality 
of  the  iron  and  the  induction  is  fixed,  and  is  independent  of  all 
other  factors.  The  power  consumed  by  the  choking  coil  will  be 
^E  .  C  cosX  or  in  this  case  £E .  C  sim£.  But  the  loss  of  energy 
was  seen  to  be  h.n.v  at  the  given  periodicity.  Further,  we 
have 

E=p.3>.T=27r.n.B.s.T 

and  H I  =  4?r .  C  .  T  or   C 


Hence         £E .  C 
But  this  leads  to 


2TT.n.B.s.  T.H.I 

8-jr.T 


47TJT 

=  $(n.B.H.v). 


(n .  B .  H .  v)  sin  -\|r  =  h .  n .  v   or   sin  >|r  = 


4fe 
B.  H' 


and  hence  *jr  is  completely  determined  when  the  maximum  induction 
is  given,  since  that  gives  definitely  the  values  of  h  and  H.  The 
usual  limits  for  ^r  are  between  45°  and  30°.  As  an  example 
consider  a  choking  coil  working  at  5000  lines  per  square  cm.,  on 
reference  to  the  curves  it  will  be  seen  that  the  corresponding  value 
of  H  is  2*87  and  of  h  is  about  1800.  Hence  we  have 

4  x  1800 

Sm*  =  2^5000  =  °'5' 
which  gives  ifr  =  30°  nearly. 

Effect  of  air  gap.     In  the  last  example  we  considered  the 
case  of  a  choking  coil  with  the  magnetic  circuit  closed,  that  is  to 

Current 


Cunnent 


P.D 


Fig.  40. 

say  without  a  gap  in  it.     If  there  be  an  air  gap  in  the  circuit  the 
hysteresis  cycle  will  be  sheared  over,  and  if  we  assume  that  the 

4—2 


52  ALTERNATING   CURRENTS 

maximum  induction  is  the  same  as  in  the  last  case,  its  form  will 
be  as  shown  in  Fig.  40 ;  on  completing  the  construction  as  before 
we  again  arrive  at  the  current  curve.  Owing  to  the  magneto- 
motive force  required  for  the  gap  it  is  evident  that  the  current  will 
be  larger  and  will  approximate  much  more  closely  to  a  sine  curve ; 
and  further  it  will  be  seen  that  the  hysteretic  angle  of  lead  is 
much  reduced  and  the  current  and  flux  are  brought  much  more 
into  phase,  that  is,  the  current  lags  more  nearly  90°  on  the  pressure. 
Hence  if  it  is  required  to  minimize  the  angle  between  the  flux 
and  the  current  it  is  desirable  to  put  an  air  gap  in  the  magnetic 
circuit.  The  effect  of  an  air  gap  in  the  circuit  on  the  angle  of 
hysteretic  lead  can  be  found  as  follows.  Let  there  be  an  air  gap 
of  the  amount  g  in  the  magnetic  circuit,  then  the  equation  giving 
the  current  becomes 

(HI  +  Bg}  =-•  47rC  .  T, 

while  the  other  equations  remain  the  same  as  before.  Hence  the 
value  of  sin  -v/r  is  given  by 

4>h.v.n.T 

n7£~s7f  (Bg  +  HI) ' 

which  reduces  to          sin  ^!r  =  • 


showing  that  when  the  form  of  the  circuit  and  the  induction  are 
given  the  absolute  size  of  the  circuit  and  the  other  factors  have  no- 
influence  on  the  angle.  Suppose  that  the  circuit  considered  in  the 
last  case  has  an  air  gap  which  is  one  per  cent,  of  the  total  length> 
we  then  have 

4  x  1800  n  MA 

m*°  4680  (2-87 +  60r 

which  corresponds  to  a  value  of  ^r  equal  to  1°  24',  showing  the 
very  great  increase  in  X  consequent  on  the  presence  of  a  small  air 
gap;  this  air  gap  however  necessitates  a  much  larger  current  as 
is  evident  from  the  figure. 

Vector  Figure.  The  vector  representation  of  the  choking 
coil  taking  into  account  the  presence  of  hysteresis  is  given  in  Fig.  41. 
The  pressure  vectors  OE  and  its  negative  OEl  are  still  at  right 
angles  to  the  flux  vector,  but  the  vectors  representing  the  current 
and  magnetomotive  force  lead  the  flux  by  the  angle  of  hysteretic 
lead.  The  loss  of  pressure  due  to  resistance  can  be  shown  as 
follows.  If  R  be  the  ohmic  resistance  of  the  coil,  the  maximum 
drop  of  pressure  will  be  given  by  the  product  CR.  This  pressure 
is  evidently  in  phase  with  OG  which  represent  the  current,  and 
hence  if  the  vector  OD  be  cut  off  from  OC  of  such  a  length  as  to 
represent  the  drop  on  the  scale  selected  for  the  pressures,  the 
impressed  pressure  will  have  to  supply  this  as  well  as  equilibrate 


THE   CHOKING   COIL   AND   TRANSFORMER 


53 


\  hence  it  will  be  given  by  the  diagonal  OF  of  the  parallelogram. 
Thus  the  current  and  pressure  are  brought  more  into  phase  which 


Fig.  41. 

must  be  the  case  since  more  power  has  to  be  accounted  for  owing 
to  the  loss  of  energy  in  the  resistance  of  the  coil. 

Eddy  current  loss.  In  deriving  the  current  from  the  cyclic 
curve  the  hysteresis  loss  was  alone  considered.  In  addition  to 
this  there  is  necessarily  a  loss  due  to  the  production  of  eddy 
currents  in  the  iron  core.  The  following  experiment  will  show 
that  such  currents  must  to  some  extent  be  taken  into  account. 


>»'  1'4 

^ 

^ 

\+* 

L 

^ 

^ 

^ 

+^- 

^ 

*^ 

^ 

^ 

'         102V30405060708C30100 
PerLods  per  second. 

Fig.  42. 

A  choking  coil  was  fed  with  alternating  currents  at  different 
periodicities  and  different  pressures,  but  the  two  quantities  were 
so  adjusted  that  the  pressure  was  proportional  to  the  periodicity 
and  hence  the  maximum  induction  in  the  iron  was  always  the 
same,  the  ohmic  drop  being  in  this  case  negligible.  The  loss  of 


54  ALTERNATING   CURRENTS 

energy  was  measured  by  means  of  a  wattmeter  in  each  case  and 
from  the  results  the  loss  of  energy  per  cycle  in  joules  was  found 
as  a  function  of  the  periodicity,  the  result  being  shown  in  Fig.  42. 
If  the  loss  were  only  due  to  hysteresis  it  would  be  a  constant,  but 
it  will  be  seen  that  it  increases  in  proportion  to  the  periodicity, 
showing  that  the  increase  is  due  to  eddy  currents  induced  in  the 
iron.  The  increase  in  the  loss  due  to  eddies  at  a  periodicity  of 
100  will  be  seen  to  be  about  30°/0.  This  is  very  far  in  excess 
of  what  occurs  in  practice  since  the  induction  used  in  the  experi- 
ment was  very  much  higher  than  is  usually  used,  and  the  loss  in 
eddies  is  proportional  to  the  square  of  the  induction;  it  will  be 
noted  that  the  intercept  on  the  vertical  axis  gives  the  constant 
hysteretic  loss. 

Examples.  As  examples  consider  the  following  cases.  A 
choking  coil  has  an  iron  core  whose  section  is  10  by  10  centimetres 
and  the  mean  axial  length  is  70  centimetres.  It  has  113  turns  on 
it  and  is  to  work  on  a  circuit  for  which  the  periods  are  about  83 
or  for  which  %7rn  is  500.  The  maximum  induction  to  be  used  is 
5000,  and  it  is  required  to  find  the  pressure  and  current  taken. 
The  maximum  value  of  the  induced  E.M.F.  being  given  by 
E  =  p<&T,  in  this  case  we  have 

500  x  5000  x  100  x  113 
E  =  —  -  =  282  volts. 

Hence  if  the  resistance  of  the  coil  be  fairly  small,  the  corresponding 
virtual  terminal  pressure  will  be  nearly  200  volts.  Since  the 
maximum  induction  is  5000  a  reference  to  the  curve  on  p.  47  will 
show  that  the  corresponding  //  is  nearly  2 '9,  hence  the  magneto- 
motive force  is  2'9  x  70  or  203.  To  find  the  maximum  value  of 
the  magnetising  current  in  amperes  we  then  have 


203, 

J.V7 

which  gives  C  =  T43. 

Hence  its  virtual  value  is  I'O  ampere.  The  current  will  be 
completely  known  when  we  find  its  hysteretic  component.  At 
B  =  5000  a  reference  to  the  curve  on  p.  50  will  show  that  the 
loss  is  1800  ergs  per  c.c.  per  cycle,  and  hence  in  this  case  the  loss 
in  watts  is 

1800  x  83  x  7000 

--TOT-  ^   105. 

Hence  the  hysteretic  component  of  the  current  is  0'52  ampere 
and  the  angle  of  hysteretic  advance  is  such  that  its  cosine  is  0'52 
or  is  about  59°. 

Suppose  that  it  is  required  to  find  the  size  of  a  choking  coil 
that  is  to  absorb  100  volts  and  permit  10  amperes  to  pass  at  the 


THE   CHOKING   COIL   AND   TRANSFORMER  55 

same  periods  as  the  last  case.  We  will  take  the  maximum  in- 
duction as  8000  for  which  the  corresponding  H  is  about  5*5,  and  a 
provisional  mean  length  of  80  cm.  Hence  to  obtain  the  number 
of  turns  we  have 

4-7T 

|^. 14.^  =  80x5-5, 

since  the  maximum  current  corresponding  to  10  virtual  amperes 
is  about  14.  This  leads  to  T  =  25.  To  find  the  corresponding 
section  for  a  maximum  pressure  of  141  volts  we  have 

141  x  108  =  500  x  8000  x  s  x  25, 

which  leads  to  5  =  141  or  a  square  section  of  11 '85  cm.  in  the  side. 
The  loss  of  energy  per  cm.  per  cycle  will  from  the  curve  be 
found  to  be  4000,  hence  the  watts  absorbed  are 

4000  x  141  x  80  x  83 

— z— —  -  or  nearly  370, 

corresponding  to  a  current  of  3'7  amperes.  The  phase  angle  is 
given  by  cos  X  =  0'37  or  is  X  =  68°.  It  may  be  noted  that  with  the 
given  induction  pressure  and  current,  the  volume  of  the  iron  must 
be  constant,  for  we  can  write 

C  =  ^VT  and  E=P-B'S'T- 

IO.H.B   , 
Hence  E  .  C  =  -  -  .Is. 

4.7T 

But  all  the  quantities  on  both  sides  are  fixed  by  the  conditions 
of  the  case,  hence  the  volume,  and  therefore  the  loss  of  power, 
must  be  the  same  for  all  arrangements  of  the  core.  To  lessen  the 
size  it  would  be  necessary  to  increase  the  induction,  but  if  this  be 
much  further  increased  the  assumption  that  the  maximum  current 
is  the  same  in  the  actual  curve  and  its  equivalent  sine  will  no 
longer  hold  good,  and  hence  the  best  arrangement  could  only  be 
settled  by  experiment. 

The  presence  of  an  air  gap  will  enable  the  choking  coil  to 
be  designed  of  much  smaller  dimensions  and  with  smaller  losses. 
Suppose  the  iron  circuit  has  a  section  of  60  sq.  cm.,  and  a  length 
of  60  cm.,  while  the  induction  is  5000  and  the  periods  are  83,  giving 
p  —  2?r7i  =  500  nearly.  If  the  pressure  and  current  are  to  be  as 
before  100  volts  and  10  amperes,  to  find  the  turns  we  still  have 

108  x  141  =  500  x  5000  x  60  x  T, 
which  lead  to  T  =  94. 

Let  us  suppose  that  an  air  gap  of  g  cm.  be  made  in  the  iron 
circuit,  its  value  must  be  given  by 


56  ALTERNATING   CURRENTS 

this  leads  to 

(2-9  x  60)  +  50(%  =  ^  x  14-1  x  94  or  g  =  0'29  cm. 

Hence  the  air  gap  enables  a  much  smaller  coil  to  be  used,  and 
the  loss  in  watts  will  now  be 

1800  x  60  x  60  x  83 

T6i- 

or  a  considerable  reduction  on  the  value  with  an  all-iron  circuit. 
The  in-phase  current  being  0*54  ampere  the  phase  angle  is  given 
by  cos  X  =  0*54  or  \  =  87°  nearly. 

Condition  of  maximum  phase  angle*.  For  the  sake  of 
simplicity  the  diminution  of  phase  angle  between  current  and 
pressure  which  is  due  to  resistance  considered  on  p.  52  has  been 
neglected  in  these  examples.  In  the  case  where  very  large  phase 
angles  are  required,  it  can  readily  be  shown  that  the  condition  for 
maximum  phase  angle  is  given  when  the  ohmic  loss  in  the  choking 
coil  is  equal  to  the  hysteretic  loss.  Let  WL  be  the  latter,  at 
constant  applied  pressure  it  will  be,  as  seen,  nearly  constant  ;  the 
ohmic  loss  will  be  ffiR,  and  hence  the  total  loss  is  WL  +  ^R. 
If  £Q  is  the  constant  applied  pressure,  we  must  have  the  apparent 
power  as  given  by  Sffi,  and  hence  we  have  £J@  cos  X  =  WL  +  ffiR. 
Thus 

WL  +  ^R 

cosx=     $y     ' 

For  a  definite  iron  core  with  an  air  gap,  the  value  of  ffi  depends 
principally,  as  we  have  just  seen,  on  the  air  gap  provided,  and 
hence  to  find  the  best  air  gap  to  give  minimum  value  of  cos  X 
this  expression  must  have  its  differential  coefficient  with  respect 
to  ^  equated  to  zero.  This  leads  to 


or  to  WL  = 

as  stated  above. 

The  transformer.  Ideal  case.  If  a  second  circuit  be 
wound  on  the  core  of  a  choking  coil  the  flux  will  pass  through 
this  also  and  hence  an  K.M.F.  will  be  induced  in  it.  This  E.M.F. 
can  be  used  to  supply  a  current,  and  in  such  a  case  the  apparatus 
is  called  a  transformer.  The  two  circuits  are  then  distinguished 
as  the  primary  and  secondary  circuits.  We  will  first  consider  that 
the  flux  through  both  coils  is  the  same,  this  is  never  in  practice 
the  case  but  in  well  designed  transformers  it  is  nearly  true,  and 
we  shall  see  that  in  most  cases  this  condition  must  be  as  nearly 
fulfilled  as  possible  In  view  of  what  has  already  been  established 

*  Professor  B.  Hopkinson. 


THE    CHOKING   COIL   AND   TRANSFORMER 


57 


we  can  at  once  proceed  to  draw  the  complete  vector  representation 
of  this  case.  Let  the  turns  on  the  coils  be  respectively  Tj  and  T2; 
the  reluctance  of  the  core  p  ;  Rp  and  R8  the  resistances  of  the  coils, 
4>  any  assumed  maximum  flux  in  the  core.  Suitable  scales  must 
be  selected  for  pressures,  currents,  magnetomotive  forces  and  flux. 
Draw  any  vector  OF  (Fig.  43)  to  represent  the  flux  4>.  From 


Fig.  43. 

the  properties  of  the  core  we  can  find  the  current  required  and  the 
angle  of  hysteretic  lead  as  in  the  previous  case  and  thus  derive  the 
vector  OM  in  direction  and  magnitude  which  will  give  the  magneto- 
motive force  required  for  this  maximum  flux.  The  two  maximum 
induced  E.M.F.S  will  be  in  the  direction  OE,  at  90°  to  OF  and  the 
lengths  of  the  vectors  representing  them  will  be  p  .  <X>  .  T^  and 
p.<b.Tz.  These  are  shown  at  OEl  and  OEZ. 

Let  us  assume  that  the  circuit  on  which  the  secondary  is 
working  is  non-inductive,  and  that  the  external  resistance  is  R. 
Then  the  maximum  current  will  be  represented  by  a  vector  in  the 
direction  of  OE2  with  the  length 

=  OC2. 


This  current  will  produce  a  magnetomotive  force  of  the  amount 


58 


ALTERNATING   CURRENTS 


which  will  be  represented  by  the  vector  OM2  in  phase  with  0(72. 
Hence  while  the  core  actually  requires  the  magnetomotive  force 
OM,  the  secondary  alone  produces  one  of  amount  OM2.  It  follows 
that  the  primary  must  produce  one  obtained  as  shown  by 
means  of  the  parallelogram  OM2MMl}  or  will  be  given  by 
the  vector  OMlf  The  corresponding  maximum  current  in  the 
primary  will  be  given  by  the  equation  OM^  =  Al  =  4?r.  T^ .  d. 
From  this  we  get  the  length  OCl  of  the  primary  current  vector, 


Fig.  44. 


Fig.  45. 


its  phase  being  the  same  as  OMlt  The  flow  of  this  current  will 
produce  a  drop  of  pressure  the  maximum  value  of  which  is  R.OCl} 
the  phase  being  the  same  as  that  of  the  primary  current ;  this  is 
shown  by  the  vector  OES.  Now  the  applied  pressure  will  have  to 
supply  this  drop  and  also  equilibrate  the  pressure  given  by  the 
vector  OE1.  The  vector  OE5  being  taken  as  OEl  reversed  it  will 
be  seen  that  from  the  parallelogram  OESVE5  we  derive  the 
vector  0  V  giving  the  maximum  primary  terminal  pressure.  The 
terminal  secondary  pressure  will  be  found  thus ;  the  drop  will  be 


THE   CHOKING   COIL   AND   TRANSFORMER  59 

given  by  Rs .  C2  and  will  be  represented  by  the  vector  OE4  in  phase 
with  0(72,  hence  the  terminal  pressure  will  be  the  difference 
between  OE.2  and  OE4  or  will  be  OE. 

A  transformer  is  generally  used  for  the  purpose  of  supplying 
apparatus  at  some  pressure  differing  from  that  existing  between 
the  supply  mains  available,  and  very  often  utilizes  the  high 
pressure  from  such  mains  to  produce  a  lower  one  for  the  apparatus. 
In  such  cases  the  pressure  on  the  supply  mains  is  usually  of 
constant  virtual  value  and  it  is  desired  to  keep  the  pressure  on 
the  terminals  of  the  secondary  as  nearly  constant  as  possible. 
The  ratio  between  the  primary  and  secondary  pressures  is  called 
the  Transformation  ratio.  In  the  figure  we  have  just  derived  this 
ratio  will  be  that  of  the  lines  0V  and  OE.  If  no  ohmic  drops 
existed  in  either  of  the  circuits  the  vectors  OE3  and  OE4  would 
be  non-existent  and  the  ratio  of  transformation  would  then  be 
that  of  the  lines  OEl  and  OE2.  But  this  is  evidently  the  ratio  of 
the  turns  in  the  coils  or  T^T^  Hence  the  nearer  we  can  approxi- 
mate to  zero  resistance  in  the  coils  the  more  constant  will  be  the 
ratio  of  transformation.  It  follows  that  in  an  actual  transformer 
where  the  ratio  of  the  applied  primary  pressure  to  the  terminal 
secondary  pressure  is  very  nearly  constant,  the  vectors  OE3  and 
OE4  are  very  small  compared  with  the  other  pressure  vectors  and 
thus  as  a  first  approximation  we  can  take  the  vectors  OEl  and  0V 
as  the  same  in  length.  Up  to  the  present  we  have  arbitrarily 
assumed  the  value  of  OF  but  we  see  that  we  can  nearly  take  it  as 
being  given  by  the  relation  OV=p.3>.Tl.  This  being  deter- 
mined the  rest  of  the  construction  follows  as  described.  We  also 
see  that  in  such  a  case  all  the  vectors  other  than  those  connected 
with  the  flux  are  closely  in  co-phase  or  anti-phase  and  that  the 
flux  vectors  are  nearly  at  90°  to  the  others. 

In  Fig.  44  is  given  a  diagram  for  another  current  larger  than 
the  last  one.  The  diagram  for  no  load  is  shown  in  Fig.  45.  It 
will  be  seen  that  the  effect  of  loading  the  transformer  is  to  bring 
the  current  and  pressure  in  the  primary  more  and  more  into  phase. 

Regulation.  The  variation  of  the  transformation  ratio  in 
a  transformer  is  a  measure  of  its  regulating  properties.  The  fall 
in  pressure  on  full  load  is  a  small  quantity,  in  general  too  small 
to  be  satisfactorily  measured  directly  by  means  of  the  difference 
between  the  readings  of  the  no  load  and  full  load  pressures  on  the 
secondary  terminals.  The  following  method  will  enable  this  point 
to  be  tested.  Two  transformers  of  the  same  type  are  joined  as 
shown  in  Fig.  46  with  their  primaries  in  parallel  on  the  supply 
mains  and  their  secondaries  arranged  in  series  with  a  low  reading 
voltmeter  V,  but  in  such  a  direction  that  the  secondary  pressures 
oppose,  one  of  the  secondaries  can  be  loaded  by  means  of  a  resis- 
tance and  the  current  taken  measured  by  the  ammeter  A.  The 


60 


ALTERNATING   CURRENTS 


reading  of  the  voltmeter  in  this  case  evidently  will  be  the  drop 
corresponding  to  the  load  being  carried  since  the  E.M.F.  of  the 


AMJUUU       UjLJUJUL 


Fig.  46. 

unloaded  one  is  a  constant,  or  if  the  pressure  on  the  mains  should 
vary  slightly  it  will  affect  both  transformers  in  the  same  way. 
It  should  be  noted  that  the  test  cannot  be  very  accurate  since  as 
will  be  seen  later  on  the  phase  of  the  E.M.F.  of  the  loaded  one  and 
that  of  the  E.M.F.  of  the  other  are  not  the  same,  and  hence  the 
reading  of  the  voltmeter  will  give  the  vector  difference  of  the 
pressures  and  not  the  actual  difference  between  the  no  load 
pressure  and  the  loaded  one. 

Case  of  a  phase  angle  in  secondary ;   lagging  current. 

We  will  now  take  the  case  where  the  circuit  of  the  secondary 
is  inductive.  The  flux  vector  OF  in  Fig.  47  is  taken  of  the  same 
length  as  before  and  the  induced  E.M.F.  and  resultant  magneto- 
motive force  vectors  are  drawn  as  in  the  last  case.  On  the 
vector  of  the  secondary  E.M.F.,  OE^,  a  semicircle  is  drawn  and 
the  angle  E2OC2  taken  equal  to  the  angle  whose  tangent  is  the 
reactance  of  the  secondary  divided  by  its  total  resistance.  Then 
OE6  will  be  the  effective  E.M.F.  in  the  secondary  or  that  required  for 
the  resistances  internal  and  external,  while  E2E6  is  the  back  E.M.F.  or 
that  required  for  the  self-induction  of  the  same.  Thus  the  current 
in  the  secondary  will  be  represented  by  a  vector  in  the  direction 

/\  TJ1~ 

of  OES  with  the  length  „      *    .     The  corresponding  magnetomo- 
-Ks  -f-  jft 

tive  force  will  be  given  by  OM^  which  is  as  before  47r.C2.l72. 
The  primary  magnetomotive  force  is  obtained  by  drawing  the 
parallelogram  OM^MM^  since  it  must  be  such  as  to  give  with  OM2 
the  resultant  OM.  In  the  same  way  as  in  the  previous  case  we 
derive  the  primary  current  OCl  and  from  it  the  drop  in  pressure  due 
to  the  resistance  of  the  primary,  OES,  and  by  combining  this  with 
the  reversed  induced  primary  E.M.F.,  OE5,  we  get  the  primary 
impressed  pressure  0  V.  The  secondary  terminal  pressure  will  be 


THE   CHOKING  COIL  AND   TRANSFORMER 


61 


found  as  follows.  Cut  off  the  vector  OE4  from  the  secondary 
effective  pressure,  it  being  equal  to  CZR8  and  join  E2E4.  Then 
E2E4  is  the  secondary  pressure.  Its  proper  angular  position  will 


M, 


Fig.  47. 


Fig.  48. 


be  got  by  drawing  OE  equal  and  parallel  to  E2E4,  so  that  the 
secondary  angle  of  lag  for  the  load  is  E4OC2.  It  will  be  seen  that 
the  angle  of  lag  in  the  secondary  is  as  it  were  transferred  to  the 
primary,  and  further  that  for  the  same  secondary  power  the 
regulation  is  worse,  or  the  ratio  of  OE  to  0  V  is  less  than  in  the 
non-inductive  case. 

Leading  current.  If  the  secondary  load  be  such  that  the 
current  leads  the  pressure  we  can  easily  draw  the  vector  diagram 
in  the  same  way.  This  is  shown  in  Fig.  48.  It  is  not  necessary 
to  follow  the  construction  in  detail,  since  the  construction  is  the 
same  as  in  the  last  case. 

Leakage.  Up  to  the  present  we  have  assumed  that  the  flux 
of  magnetism  is  the  same  through  both  the  primary  and  secondary 
coils.  This  can  never  be  the  case  since  the  two  coils  would  then 
have  to  occupy  identically  the  same  position.  As  an  extreme 
example  consider  the  arrangement  shown  in  Figs.  49  and  50.  The 
magnetomotive  force  produced  by  the  primary  coil  will  tend  to  send 
a  large  flux  through  the  iron  core  and  in  addition  another  flux 
through  the  various  paths  provided  by  the  surrounding  air  which 
is  called  the  leakage  field  or  leakage  flux.  Hence  at  no  load  on 


62 


ALTERNATING    CURRENTS 


the  secondary  the  flux  will  be  as  shown  in  Fig.  49,  where  the 
major  part  of  the  flux  passes  down  the  iron  but  a  small  part  will 


Fig.  49. 


Fig.  50. 


flow  outside  it,  the  maximum  of  the  core  flux  will  always  be  less 
than  the  maximum  of  the  flux  that  passes  through  the  primary 
coil  and  the  ratio  of  the  two  is  determined  solely  by  the  relative 
reluctances  of  the  core  and  the  air  leakage  paths.  When  a  current 


Fig.  51. 


THE   CHOKING  COIL  AND  TRANSFORMER  63 

is  flowing  in  the  secondary  the  magnetomotive  force  due  to  it  will 
almost  directly  oppose  the  primary  magnetomotive  force  and 
hence  tend  to  drive  the  flux  backwards ;  that  is,  the  core  flux  will 
no  longer  all  pass  through  the  secondary  but  some  of  it  will  be 
forced  out  into  additional  leakage  paths  as  shown  in  Fig.  50 

In  an  actual  transformer,  such  as  is  shown  in  Fig.  51,  the 
two  coils  are  not  wound  on  distinct  parts  of  the  core  but,  to  avoid 
this  leakage  effect,  are  closely  interleaved  or  otherwise  arranged 
so  that  the  opposing  magnetic  effects  of  the  two  currents  may  as 
nearly  as  possible  act  together  at  all  points  of  the  core.  But 
however  small  the  subdivision  may  be  made,  the  presence  of  this 
leakage  of  flux  cannot  be  avoided.  Consider  the  case  indicated  in 
Fig.  52  where  the  coils  are  supposed  to  be  interleaved,  and  a  few 


Fig.  52. 

successive  portions  are  shown.  When  the  current  in  the  primary 
sections,  P,  is  a  positive  maximum  as  shown  by  the  dots  and 
crosses  on  the  sections,  a  dot  expressing  that  the  current  is  flowing 
upwards  in  the  wires,  a  cross  that  it  is  flowing  downwards,  the 
secondary  currents  will  be  practically  at  their  negative  maximum, 
and  the  dots  and  crosses  for  them  will  be  as  shown.  Each  of  these 
sets  of  small  coils  will  evidently  produce  small  whirls  of  magnetism 
round  themselves  as  shown,  and  these  whirls  will  pass  almost 
entirely  through  the  air  space  near  the  coils.  It  follows  that  the 
reluctance  of  the  path  through  which  each  little  whirl  of  flux 
passes  will  be  of  practically  constant  amount  and  that  the  amount 
of  flux  in  each  whirl  will  be  proportional  to  the  current  in  the 
section  that  is  producing  it.  The  leakage  fluxes  from  all  these 
separate  small  whirls  in  either  coil  will  superpose  themselves  on 
any  main  flux  that  may  be  existent  owing  to  the  combined  effect 
of  all  the  coils  in  a  similar  way  to  the  case  first  considered.  That 
is  to  say,  the  useful  flux  that  gets  into  the  secondary  will  be 
diminished  by  the  presence  of  these  little  fluxes,  while  the  flux 
that  gets  cut  by  the  primary  will  be  increased  by  its  own  leakage. 
It  is  evident,  however,  from  the  figure  that  the  average  value  of 
the  core  flux  is  not  affected,  each  little  whirl  of  leakage  flux  in 
successive  sections  of  the  coils  will  flow  in  opposite  directions  as 


64 


ALTERNATING  CURRENTS 


shown,  and  thus  we  can  take  the  core  flux  as  being  practically 
constant  along  its  length.  Hence  the  calculation  of  the  current 
required  by  the  core  for  a  definite  flux,  and  the  evaluation  of  the 
angle  of  hysteretic  lead  remain  the  same  as  before.  Let  us  denote 
by  3>  this  core  flux,  by  <J>2  the  corresponding  nett  secondary  flux 
after  the  leakage  flux  has  been  taken  away,  and  by  <E>!  the  value 
of  the  primary  flux  after  its  leakages  have  been  added. 

The  ratio  of  the  maximum  values  of  the  three  fluxes  will  now 
depend  not  only  on  the  reluctances  of  the  core  and  the  two  leakage 
paths  but  also  on  the  phase  relations  of  the  magnetomotive  forces  and 
their  magnitudes.  Let  such  a  leaky  transformer  be  working  with 
its  secondary  on  a  non-inductive  load,  and  let  Of  in  Fig.  53  denote 


Fig.  53. 

the  flux  that  is  passing  through  the  secondary.  The  E.M.F.  induced 
in  that  coil  will  be  given  by  the  vector  OE2  lagging  90°  after  Of 
and  of  the  magnitude  OE2  equal  to  pT2<&2.  If  the  external  resist- 
ance of  the  secondary  circuit  be  R  ohms,  its  total  resistance  is 
Rg  +  R  and  hence  the  current  will  be  OE2  divided  by  this,  let  it  be 
given  by  OCZ.  This  current  will  produce  a  magnetomotive  force, 
OM2,  in  the  direction  of  the  vector  OC2  and  of  the  amount 

Az  =  ——.  T2.  OG2, 

which  acting  on  the  reluctance  of  the  secondary  leakage  paths  will 
give  a  leakage  flux  of  the  amount  <£>S2  and  denoted  by  0-*jr2,  in  phase 
with  the  secondary  current  and  of  amount  proportional  to  the 
secondary  current.  Then  since  the  core  flux  must  be  such  as  to 
give  Of  as  the  resultant  flux  in  the  secondary  when  it  is  combined 


THE   CHOKING   COIL   AND   TRANSFORMER  65 

with  the  secondary  leakage  flux  ^>K,  it  will  be  given  in  maximum 
value  and  phase  by  the  vector  0$.  The  reluctance  of  the  core 
and  its  angle  of  hysteretic  advance  being  known  we  can  draw 
the  vector  OM  to  represent  the  resultant  magnetomotive  force, 
the  angle  of  hysteretic  lead  being  M0$.  Now  this  magnetomotive 
force  is  the  resultant  of  the  two  due  respectively  to  the  primary 
and  secondary  currents,  hence  as  before  the  primary  magnetomotive 
force  will  be  given  by  the  vector  OM^  found  by  drawing  the 
parallelogram  OMJfMi.  The  primary  current  will  be  found  in 
the  way  described  before  and  can  be  represented  by  the  vector  OClf 
But  the  primary  magnetomotive  force  also  acts  on  the  primary  leak- 
age paths  and  will  produce  a  flux  through  them  proportional  to  OMl 
and  of  amount  equal  to  this  M.M.F.  divided  by  the  reluctance  of 
those  paths.  Let  this  flux  be  <X>gl  and  be  given  by  Ofa  in  phase 
with  OMj.  and  proportional  to  Cj.  By  drawing  the  parallelogram 
shown  we  get  the  vector  OF  which  will  represent  the  maximum 
primary  flux  4v  The  primary  induced  E.M.F.  will  be  given  by  the 
vector  OEl  at  right  angles  to  OF  and  of  a  length  given  by  p.T-^.^. 
The  pressure  lost  in  resistance  in  the  primary,  OES,  will  be  in 
phase  with  OC \  and  of  amount  OCi  multiplied  into  the  primary 
resistance  Rp.  Hence  the  primary  impressed  pressure  will  be  the 
resultant  of  this  and  OE^  reversed  (that  is  OE5)  or  will  be  given 
by  0V.  The  secondary  terminal  pressure  will,  as  before,  be  found 
by  subtracting  from  OE2  the  vector  OE4  which  is  equal  to  C2 
multiplied  into  the  secondary  resistance,  and  hence  that  pressure 
will  be  given  by  OE. 

It  will  be  noticed  that  the  effect  of  leakage  is  to  produce 
a  larger  phase  difference  between  the  primary  pressure  and 
current  than  would  have  existed  with  no  leakage,  or  in  other 
words  the  presence  of  leakage  produces  the  same  effect  as  if  the 
secondary  of  a  non-leaky  transformer  were  working  on  an  inductive 
load.  Another  important  effect  produced  is  worse  regulation ;  for 
even  if  the  resistance  drops  be  neglected,  the  ratio  of  transforma- 

T 

tion  is  no  longer  given  by  the  ratio  of  the  turns,  -^  ,  but  by  the  ratio 

-*2 

^~- ~ .     But  with  a  constant  impressed  pressure  the  value  of  4>j 

^2-^2 

remains  nearly  constant  while  the  two  leakage  fluxes,  <&n  and  4>«2, 
will  increase  proportionally  to  the  currents,  hence  4>2  will  continually 
diminish  as  the  load  increases  in  the  secondary,  and  hence  the 
regulation  will  be  greatly  affected  by  leakage. 

The  student  should  repeat  this  construction  with  a  lag  in  the 
secondary  and  also  with  a  lead.  In  the  former  case  he  will  find 
that  the  ratio  of  the  two  pressures  is  still  more  affected  showing 
that  leakage  must  be  avoided  when  the  load  is  inductive.  In  the 
former  he  will  find  that  the  leading  current  tends  to  diminish  the 
evil  effects,  which  must  be  the  case  since  capacity  will  tend  to 
diminish  the  self-induction  effect  of  the  leakage. 


L. 


CHAPTER   V. 


TRANSFORMER  EQUATIONS. 


Transformer :  analytical  expression.  Leakage  equi- 
valent to  a  reactance.  If  the  graphical  methods  just  described 
of  representing  the  performance  of  a  transformer  be  applied  to  any 
practical  type,  it  will  be  found  that  they  are  not  suitable  for 
investigating  its  properties  since  the  currents  concerned  in  the 
magnetic  cycle  and  the  losses  of  pressure  in  the  coils  are  very 
small  compared  with  the  other  currents  and  pressures.  It  is 
important,  therefore,  to  find  some  more  appropriate  method  of 
dealing  with  the  question.  Since  the  currents  concerned  with  the 
transference  of  power  from  one  circuit  to  the  other  are  by  far  the 


Fig.  54. 

most  important,  let  us  at  first  neglect  the  small  magnetising 
current  altogether.  Then  if  OC2,  Fig.  54,  be  any  current  in  the 
secondary  it  will  be  accompanied  by  an  exactly  antiphased  current 
OCl  in  the  primary,  the  values  of  the  two  being  in  the  ratio  of  the 


OF    THE 

UNIVERSITY 


TRANSFORMER   EQUATIONS  67 

turns.  If  OE  be  the  secondary  induced  E.M.F.,  the  flux  corre- 
sponding will  be  given  by  Of.  It  will  follow  that  the  leakage  flux 
4>s2  due  to  the  secondary  current  will  be  given  by  (f>f  in  phase  with 
it,  and  4>«,  that  due  to  the  primary,  by  F<f>  in  phase  with  its  current, 
but  since  the  two  currents  are  exactly  antiphased,  the  points  /,  <f>, 
and  F  are  in  a  line.  We  can  thus  consider  the  vector  Ff  as  giving 
us  the  total  leakage  flux,  <&g,  of  the  transformer  and  the  vector  OF 
will  be  the  total  flux  cut  by  the  primary  wires.  This  will  induce 
an  E.M.F.  therein  given  by  OE^.  But  we  can  consider  this  E.M.F. 
as  being  made  up  of  two  components,  the  one  due  to  the  flux  Of 
the  other  due  to  the  leakage  flux,  Ff,  these  E.M.F.S  being  as  shown 
at  OEa  and  OEi.  The  former  will  be  related  to  the  secondary 
E.M.F.,  OE,  merely  in  the  ratio  of  the  turns  since  both  are  due  to 
the  flux  Of,  the  latter  will  be  proportional  to  the  current  in  the 
primary  of  the  transformer,  and  will  be  in  quadrature  therewith, 
that  is  perpendicular  to  Ff.  Hence  the  primary  pressure  must  be 
given  by  the  resultant  of  these  two  vectors  when  reversed,  or  will 
be  given  by  the  two  vectors,  —  OEa  and  08.  It  will  again  be 
readily  seen  that  the  E.M.F.  due  to  the  leakage  field  behaves  just  as 
if  the  primary  had  a  definite  reactance.  For  the  possession  of 
a  reactance  of  the  value  S  would  mean  that  with  a  current  *& 
passing  an  E.M.F.  of  the  value  S  .  ^  was  produced  in  the  primary 
and  that  E.M.F.  would  be  in  quadrature  with  the  current.  But  the 
leakage  field  is  proportional  to  the  current  and  hence  the  E.M.F.  due 
to  it  would  be  also  proportional  to  the  current  and  to  the  periods 
and  would  be  in  quadrature  with  the  Current.  Hence  at  constant 
periods  the  effect  of  the  leakage  field  is  exactly  the  same  as  that 
which  would  be  produced  if  we  imagined  the  leakage  fields  of 
both  primary  and  secondaiy  suppressed  but  that  the  primary 
possessed  a  definite  reactance.  It  follows  that,  as  far  as  these  fields 
are  concerned,  we  may  express  their  result  by  saying  that  the  ideal 
transformer  has  a  definite  reactance,  S,  at  the  given  periods  and 
pressure. 

Total  equivalent  primary  resistance.     We  can  now  see 

that  in  a  similar  way  the  effect  of  the  resistance  in  the  secondary 
can  be  transferred  to  the  primary.  For  let  the  ratio  of  the  primary 
turns  to  the  secondary  ones  be  p,  let  the  actual  ohmic  resistance 
of  the  primary  be  Rp  and  of  the  secondary  Rs.  If  a  current  of  the 
amount  ^  be  flowing  in  the  primary  it  will  produce  a  loss  of 
energy  of  the  amount  *^2.  Rp.  But  the  corresponding  current  in 
the  secondary  will  be  p  times  the  primary  current,  and  the 
loss  of  energy  will  be  pz.^z.Rg.  Hence  if  we  imagine  that  an 
extra  resistance  of  the  amount  p2  .  R8  exists  in  the  primary,  and 
the  secondary  is  devoid  of  resistance,  the  result  will  be  the  same 
as  in  the  actual  case,  or  if  the  primary  be  taken  to  have  an 
equivalent  total  resistance  of  Rp  +  Rg  .  p2  we  can  consider  the 
secondary  as  devoid  of  resistance.  Thus  instead  of  the  actual 

5—2 


68 


ALTERNATING   CURRENTS 


distribution  of  leakage  fields  and  resistances  we  can  imagine  that 
the  primary  has  a  definite  resistance  which  will  allow  for  all  the 
existent  ohmic  losses  being  considered  as  due  to  the  passage  of  the 
primary  current  through  this  resistance,  and  a  definite  reactance 
which  will  produce  an  effect  exactly  the  same  as  the  actually  existent 
leakage  fields.  Hence  we  may  consider  that  the  primary  has  an 
impedance  corresponding  to  these  two  quantities,  and  at  fixed 
alternations  this  impedance  will  be  a  definite  quantity.  Thus  if 
we  denote  the  equivalent  resistance  by  R  and  the  reactance 
equivalent  to  the  leakage  E.M.F.s  by  S,  the  impedance  will  be 


/  = 


+ 


and  if  OC  (Fig.  55)  represent  any  value  of  the  primary  current  it 
will  always  be  accompanied  by  two  E.M.F.s,  the  one,  ^R,  in  phase 


with  it,  the  other,  Sffl,  in  quadrature,  the  resultant  being  <@I  and 
inclined  to  the  current  vector  at  an  angle  a  whose  tangent  is  8/R. 

Short-circuit  test.  The  value  of  this  apparent  impedance  of 
a  transformer  can  be  found  as  follows.  Short-circuit  the  secondary 
by  means  of  a  thick  wire  and  apply  an  alternating  pressure  of  the 
proper  periodicity  to  the  primary,  observing  the  primary  current 
and  pressure,  and  the  power  taken  by  the  same.  Let  these  be 
respectively  ^,  S8t  and  W8,  then  W8  —  ££@s  cos  X  where  A  is  the 
angle  between  the  pressure  and  the  current.  But  from  the  circum- 
stances that  the  secondary  is  short-circuited,  all  the  losses  that 
occur  are  those  incident  to  the  circulation  of  the  current  against 
the  resistances  of  the  two  coils  together  with  a  loss  in  the  core. 
But  the  latter  is  extremely  small  since  the  pressure  necessary  to 
circulate  even  full  load  primary  current  will  necessitate  a  mere 
fraction  of  the  full  load  pressure,  and  hence  the  cycle  of  flux  in 
the  core  will  be  of  such  a  small  magnitude  that  the  corresponding 
core  losses  will  be  negligible.  Thus  the  sole  loss  of  energy  is,  as 


TRANSFORMER   EQUATIONS  69 

said,  that  due  to  resistances  in  the  coils.  The  leakage  field 
produced,  will  be  of  necessity  that  corresponding  to  the  actually 
existent  primary  current  and  is,  like  the  resistance  losses,  dependent 
solely  on  that  current ;  hence  the  resistance  and  reactance  of  the 
transformer  are  the  same  in  this  test  as  they  would  be  when  used 
in  the  ordinary  manner  when  the  same  current  is  flowing.  For  all 
ordinary  ranges  of  currents  it  will  be  found  that  cos  A,  remains 
constant. 

Now    we   must   have   ^R  —  Ws,  where   R   is   the   required 
& 

equivalent  resistance,  and  -^  =  /,  where  I  is  the  impedance,  from 

W8 

which  the  reactance  can  be  calculated  from  the  expression 

I*  =  S*  +  R\ 

In  a  particular  transformer  of  somewhat  old  type  it  was  found 
that  to  circulate  a  current  of  0'8  ampere  in  the  primary  when  the 
secondary  was  short  circuited  a  pressure  of  86  volts  was  required, 
and  the  power  taken  was  32'7  watts.  From  this  we  readily  derive 
that  the  equivalent  resistance  is  51  ohms,  the  impedance  is  107 
and  the  reactance  is  94.  Hence  whenever  a  current  ^  is  flowing 
into  the  transformer's  primary  it  will  necessitate  the  primary 
supplying  a  pressure  of  51 .  ^  volts  in  phase  with  the  current  for 
the  resistance,  and  also  a  pressure  of  94.^  volts  leading  the  current 
by  a  right  angle.  These  pressures  are  supplied  by  the  source  of 
potential  difference,  and  what  is  left  will  be  available  for  other 
purposes  as  for  example  to  supply  pressure  to  the  secondary 
circuit,  the  secondary  coil  being  now  considered  as  devoid  of  both 
resistance  and  leakage. 

Expression  for  the  secondary  pressure.  The  whole  treat- 
ment of  the  problem,  neglecting  the  magnetising  current,  can  now 
be  referred  to  the  primary  side,  and  since  with  sinusoidal  quantities 
the  maximum  is  always  \/2  of  the  virtual  value,  the  latter  can  be 
taken  instead  of  the  former  in  drawing  up  a  diagram.  Let  OC2 
(Fig.  56)  be  the  direction  of  the  secondary  current ;  its  phase  angle, 
X,  relative  to  the  secondary  potential  difference  being  MOC2. 
The  corresponding  current  in  the  primary  will  be  less  than  this  in 
proportion  to  the  ratio  of  the  turns  but  in  exact  antiphase,  let  it 
be  given  by  OC^.  Produce  the  line  OM  upwards  to  Y  and  draw 
the  perpendicular  OX.  Then  the  nett  pressure  in  the  primary 
that  is  requisite  to  produce  the  actual  terminal  pressure  in  the 
secondary  will  be  on  OF  as  shown  at  OE.  The  actual  secondary 
pressure  will  be  l/pih  of  this.  The  current  will  be  accompanied 
by  the  two  E.M.F.S  OR  and  OS  as  before,  the  resultant  of  these 
"being  OL.  To  find  the  position  of  the  primary  terminal  pressure 
vector  a  circle  should  be  drawn  with  its  value  as  radius,  and  the 
line  LV  drawn  from  L  parallel  to  OF.  The  vector  OF  will  then 
represent  the  primary  pressure  in  magnitude  and  phase.  Hence 


70 


ALTERNATING   CURRENTS 


the  line  VE  being  drawn  parallel  to  OL  will  give  the  value  of  the 
part  of  the  primary  pressure  that  is  available  for  direct  transforma- 
tion to  the  terminals  of  the  secondary,  in  other  words  p  times  that 
terminal  pressure. 

From  these  considerations  we  can  readily  deduce  an  important 
relation  between  the  primary  pressure  and  the  constants  of  the 


transformer.  Take  the  projections  of  each  E.M.F.  on  the  two  lines 
OX  and  OF.  If  ^  be  the  primary  current  required  to  equilibrate 
the  secondary  one,  the  latter  will  be  p^;  the  angle  between  this 
and  the  line  OF  is  the  lag  in  the  secondary  circuit  or  X;  let  R 
and  S  be  the  resistance  and  reactance  of  the  equivalent  primary, 
and  let  <^  be  the  unknown  value  of  the  secondary  terminal  pressure, 
so  that  pS  is  the  equivalent  pressure  in  the  primary  :  further  let 
SQ  be  the  value  of  the  constant  primary  terminal  pressure.  Then 
the  two  sets  of  horizontal  (X)  and  vertical  (F)  projections  will 
have  the  following  values  : 

X  =  ^  (R  sin  X  -  S  cos  X),     F  =  pS  +  ^  (  R  .  cos  X  +  8  .  sin  X), 


also 


+  F 


The  above  expression  can  also  be  proved  as  follows.  Let  OC 
(Fig.  57)  be  the  direction  of  the  current  both  primary  and 
secondary,  X  the  angle  of  lag  between  the  secondary  terminal 
pressure  and  its  current,  and  therefore  between  the  primary 
current  and  the  primary  pressure  pg,  equivalent  to  the  secondary 
one,  that  is  the  line  OE.  Let  ER  be  the  value  of  the  ohmic  drop, 


TRANSFORMER   EQUATIONS  71 

<$.R,  in  the  primary  as  found  from  the  short-circuit  experiment, 
and  R  V  the  corresponding  value  of  the  reactance  E.M.F.  <$.  S.  Then 
the  primary  applied  pressure  will  be  the  resultant  of  these  as 
shown  at  OF.  Draw  the  perpendicular  RA  from  R  on  OE 


B 


Fig.  57. 


produced  and  the  perpendicular  VBD  to  the  same  from  F,  the 
point  B  being  where  this  line  meets  OE  produced,  and  the  point 
D  where  it  meets  a  line  drawn  from  R  parallel  to  OE.  Then  the 
angles  AER  and  RVD  are  each  equal  to  X,  and  since  ER  is  equal 
to  ^ .  R  and  VR  to  <@ .  S,  we  have 

/Sf.^.cosX, 


'.  sin  X)2  +  (S .  <@.  cos  \  -  ^ .  R  .  sin  X)2, 


and 

Hence 

while 

But  since 

we  get 

£*  =  (p£+<@. 

the  same  result  as  before. 

Open-circuit  test.  Up  to  the  present  nothing  has  been  said 
relative  to  the  current  required  to  produce  the  cycle  of  flux  in  the 
core,  including  the  provision  of  the  necessary  loss  of  energy 
incident  to  hysteresis,  etc.  The  magnitude  of  this  cycle,  and 
hence  that  of  the  corresponding  current,  will  depend  on  the 
pressure  induced  in  the  primary,  and  on  the  periods ;  the  latter 
will  be  constant  for  a  given  state  of  supply,  but  the  former  will 
not  be  constant.  Owing  to  resistance,  the  induced  pressure  in  the 
primary  must  always  be  less  than  the  potential  difference,  but  the 
condition  of  operation  is  such  that  the  latter  is  kept  constant, 
hence  the  cycle  of  flux  will  have  less  amplitude  the  greater  the 
current.  In  all  practical  cases  the  fall  of  pressure  is  a  small 
percentage  of  the  potential  difference,  and  we  can  very  nearly 


72  ALTERNATING   CURRENTS 

consider  the  cycle  of  flux,  and  consequently  the  current  taken 
by  the  transformer  for  the  production  of  that  flux,  as  being 
constant  at  all  loads.  Hence  if  we  can  determine  it  for  one  state 
of  operation  that  will  be  sufficient.  It  is  readily  found  for  the 
case  where  the  secondary  circuit  is  open,  and  the  determination  of 
the  corresponding  pressure,  current  and  power  in  such  a  case  is 
termed  the  open  circuit  or  no  load  test  of  the  transformer.  In  such 
a  test  it  will  readily  be  seen  that  the  current  flowing  is  very  small 
compared  with  the  full  load  current,  and  it  flows  only  against  the 
primary  resistance,  hence  the  ohmic  loss  is  in  this  case  entirely 
negligible,  and  all  the  measured  loss  can  be  considered  as  that 
incident  to  the  core  flux.  Hence  in  any  transformer  let  the 
secondary  circuit  be  open  and  measure  the  pressure  £0,  current 
^o,  and  power  W0  taken  by  the  transformer,  the  former  being  the 
normal  pressure  at  which  it  is  required  to  operate.  It  follows  that 
the  component  of  the  current  that  is  in  phase  with  the  pressure, 

W 

or   the  power  component  ffip,  is  given  by  -j?  ,  and  hence   the 

wattless  or  quadrature  component  will  be 


It  may  be  noticed  that  the  angle  of  phase  difference  between  the 

60 

current  and  pressure,  that  is  the  angle  whose  cosine  is  ^~  ,  is  the 

Wq 

complement  of  the  angle  of  hysteretic  lead.  In  the  transformer 
considered  before,  it  was  found  that  under  a  pressure  of  2000  volts 
a  current  of  0*072  ampere  flowed,  and  the  power  taken  was 
109  watts.  It  readily  follows  that  the  power  component  of 
the  current  is  0'054  ampere  while  the  wattless  current  is 
0'048  ampere,  from  which  the  angle  of  lag  readily  follows, 
being  given  by 

0'48 

or  0'8,  nearly. 


Should  it  be  difficult  to  apply  the  proper  primary  pressure  the 
observations  can  be  taken  on  the  secondary  circuit  with  the 
appropriate  pressure.  The  cycle  to  which  the  iron  is  subjected 
and  the  power  used  will  be  the  same,  but  the  observed  current 
and  pressure  must  be  reduced  to  their  equivalent  values  for  the 
primary  circuit.  In  either  case  the  core  loss  is  accompanied  by 
a  certain  ohmic  loss  in  the  coil,  but  this  will  of  necessity  be 
negligibly  small,  since  the  current  taken  is  a  very  small  fraction 
of  any  reasonable  load  current. 

Expression  for  total  current  in  primary.  The  mag- 
netising current  being  small  will  produce  practically  no  effect 
with  regard  to  fall  of  pressure  due  to  ohmic  resistance  or  in 
producing  a  leakage  flux  ;  further,  any  effects  of  such  a  current  will 


TRANSFORMER   EQUATIONS 


73 


"be  solely  referable  to  the  actual  primary  resistance  and  leakage 
and  not  to  the  equivalent  leakage  and  resistance,  and  will  hence 
produce  entirely  negligible  effects.  But  in  order  to  find  the 
actual  current  that  is  Sowing  in  the  primary,  we  must  combine 
with  the  previously  considered  power  current  the  current  that  is 
required  to  maintain  the  magnetic  cycle.  But  it  has  just  been 
shown  that  the  latter  OCm  (Fig.  58)  had  two  components,  the  one, 


Fig.  58. 


Fig.  59. 


^p,  in  phase  with  the  pressure,  the  other,  ^q,  in  quadrature  there- 
with. The  load  current  will  likewise  have  two  components  in 
the  same  directions,  namely  ^cosX  in  phase  with  the  pressure 
induced,  and  ^sinX  in  quadrature.  Hence  the  two  components 
of  the  actual  primary  current  will  be 


and  the  resultant  current  will  be  given  by 


Example.  Take  as  an  example  the  transformer  whose  con- 
stants we  have  given  and  let  us  assume  that  a  lagging  current  of 
20  amperes  is  flowing  in  the  secondary,  the  angle  of  lag  being  30° 
(Fig.  59).  The  ratio  of  transformation  was  20  to  1  so  that  the 
corresponding  primary  current  is  1  ampere.  This  means  that  the 
values  of  <@.R  is  51  volts  while  that  of  <@  .  8  is  94  volts.  Hence 
we  can  write  down  X  and  Y  as  follows  : 

Z  =  (0-5  x  51)  -  (0-86  x  94)  =  -  56, 

F  =  p<£+(0-86x  51)  +  (0-5  x  94)  =  ^+91. 

The  constant  terminal  pressure  being  2000  volts  the  following 
expression  will  give  the  value  of  £  : 


74 


ALTERNATING   CURRENTS 


This  leads  nearly  to 


pg=1903  or  £  =  95. 


Hence  under  such  circumstances  the  terminal  pressure  on  the 
secondary  will  be  about  95  volts  instead  of  100  volts,  that  would 
exist  in  the  case  of  no  resistance  or  leakage  or,  very  nearly,  on 
open  circuit.  By  taking  a  set  of  currents  at  this  angle  of  lag,  the 
curve  connecting  the  current  and  pressure  for  a  load  with  a  power 
factor  of  cos  30°  or  0'86  can  be  determined. 

It  will  be  seen  that  the  presence  of  the  additional  E.M.F.s  in 
the  primary  will  result  in  the  angle  between  the  pressure  and 
primary  current  being  altered  from  the  angle  in  the  secondary  by 
the  angle  VOE  in  Fig.  56,  and  that  this  angle,  -fy,  has  its  tangent 
given  by  X/Y.  In  most  cases  X  is  very  nearly  equal  to  the 
terminal  primary  pressure,  and  hence  we  can  nearly  write, 
tamjr  =  X/&Q.  In  this  case  X  is  56,  and  So  is  2000,  hence  tamjr 
is  nearly  0'028  or  ty  is  about  1°  40'.  Hence  the  angle  between 
the  primary  pressure  and  current  will  be  about  31°  40'.  To  find 
the  total  current  taken  we  must  refer  to  p.  72  where  it  will  be 
seen  that  the  magnetising  current  was  nearly  at  45°  to  the  pressure 
and  had  approximately  the  two  components  0'05.  Hence  the  two 
components  of  the  total  current  are  (2  cos  X  +  '05)  and  (2  sin  X-f  *05) 
or  1'05  and  1*78,  and  that  current  itself  is 


VriO  +  3'20  or  about  21  amperes. 

The  load  being  full  load  the  magnetising  current  thus  has 
practically  no  effect.  It  is  only  for  considerably  smaller  loads  that 
it  is  necessary  to  take  it  into  account. 

As  a  further  example  of  the  method,  consider  the  case  where 


Fig.  60. 

the  load  is  still  20  amperes  on  the  secondary,  but  that  circuit  is 
entirely  non-inductive,  then  the  vectors  are  as  in  Fig.  60.     We 


TRANSFORMER   EQUATIONS 


75 


can  readily  see  that  X  is  then  p£+  51  while  Y  is  -  94.  Thus  the 
terminal  secondary  pressure  is  about  97£  volts,  while  ty  is  2°  40'  or 
the  primary  angle  of  lag  is  that  amount. 

Now  take  the  case  where  the  load  is  entirely  inductive  as  in 
Fig.  61.     Here  X  is  +  51  while  Y  is  pg+  94.     Thus  the  terminal 


/>£ 


57 


Fig.  61. 

pressure  is  about  95  volts  while  the  angle  ^jr  is  1°  30'.  Since  X  is 
here  positive,  it  means  that  ty  must  be  subtracted  from  the  lag  in 
the  secondary,  or  the  nett  angle  of  lag  in  the  primary  is  88°  30'. 

Consider   lastly   the   case  of  a   load   leading  by  45°  in  the 
secondary,  Fig.  62.     From  the  figure  it  is  evident  that  the  vertical 


Fig.  62. 

components  of  R .  <$  and  S .  <$  are  now  subtracted,  while  the 
horizontal  ones  add,  hence  the  value  of  X  can  be  seen  to  be  — 101 
while  Y  is  pS—  30.  Thus  the  terminal  pressure  is  101*5  volts 


76  ALTERNATING   CURRENTS 

while  i/r  is  about  3°.  Hence  the  primary  lead  is  42°,  while  the 
secondary  pressure  is  raised  by  the  leading  current  as  we  saw  on 
p.  65.  It  will  also  be  noted  that  this  rise  in  pressure  is 
consequent  on  the  reactive  effect  of  the  primary  overpowering  its 
resistance  one,  that  is  the  projection  of  S  .*$  being  greater  than 
that  of  E  .  <$. 

By  thus  taking  various  conditions  of  loading  into  consideration 
a  full  study  of  the  action  of  the  transformer  can  be  obtained,  based 
solely  on  the  open-  and  short-circuit  tests.  When  we  come  to 
consider  the  efficiency  of  a  transformer  we  shall  see  that  the  same 
two  tests  suffice  to  determine  this  quantity  also.  They  are  thus 
of  fundamental  importance  in  the  testing  of  transformers. 

Constants  for  modern  type.  The  following  data,  referring 
to  a  good  modern  transformer,  will  give  some  idea  of  the  magnitude 
of  the  quantities  we  have  been  considering  in  such  a  case.  The 
transformer  was  of  30  kilowatt  out-put  with  a  primary  pressure  of 
2000  volts  at  60  periods  and  a  transformation  ratio  of  20  to  1. 
The  mean  length  of  the  iron  circuit  was  about  135  centimetres, 
the  cross  section  of  the  iron  about  125  square  centimetres,  and 
the  volume  about  17,000  cubic  centimetres.  The  number  of 
primary  turns  was  920.  With  the  full  load  current  of  15  amperes 
circulating  in  the  primary  the  pressure  required  on  short-circuit 
was  53'5  volts,  the  loss  being  about  500  watts,  while  the  core  loss 
was  found  to  be  about  400  watts  at  the  normal  pressure  and 
periods. 

We  will  first  find  the  induction  in  the  iron.  Since  the  pressure 
at  the  terminals  was  2000  volts  with  sine  conditions  this  cor- 
responds to  a  maximum  value  of  2830  volts.  The  periods  per 
second  being  60,  the  corresponding  value  of  p  or  2?m  is  120?r  or 
378.  But  we  know  from  p.  47  that  the  maximum  flux  in  the  core 
will  be  given  by  2830  x  108  =  378  x  920  x  <l>,  the  factor  108  being 
used  to  turn  volts  into  absolute  units  of  pressure,  hence  the  total 
core  flux  is  <3>  =  8  10  x  105  lines.  It  follows  that  since  the  cross 
section  is  125  cm.,  the  value  of  B,  the  maximum  induction  per 
square  cm.,  is  nearly  6500. 

For  a  fair  average  iron  of  the  quality  used  in  transformers  at 
this  value  of  B  the  value  of  H  will  be  about  4,  hence  if  ^0  is 
virtual  value  of  the  no  load  current  we  have 


T  being  the  turns,  I  the  length,  this  leads  to 

^  =  --  *3S  x  1QQ  -  =  0-33  ampere. 
TT  x  1-414  x  920 

Again  the  loss  in  hysteresis  for  the  same  quality  of  iron  at  the 


TRANSFORMER   EQUATIONS  77 

given  induction  would  be  about  3000  ergs  per  cubic  cm.  per  cycle, 
hence  the  loss  in  watts  due  to  hysteresis  will  be  about 

3000  x  17,000  x  60  x  10~7 
or  340  watts. 

The  difference  between  this  and  the  observed  value  will  be 
partly  due  to  eddies  in  the  core  stampings  and  partly  to  eddies 
elsewhere. 

Since  the  short  circuit  test  gave  that  the  pressure  of  53'5  volts 
was  required  to  send  a  current  of  15  amperes  it  follows  that  the 
impedance  is  35'8  ohms,  and  since  the  energy  loss  was  500  watts, 

the  equivalent  resistance  is  —^  or  2'22  ohms.     Hence  the  react- 

ance will  be  (3'582  +  2'222)*  or  2*78.  The  amount  of  flux  that 
must  leak  to  produce  this,  the  full  load  reactance,  can  be  found  as 
in  the  primary  induced  E.M.F.  by  multiplying  by  10s  and  dividing 
•by  the  value  of  2?rn  and  the  turns,  that  is  the  leaking  flux  will  be 

2'78xl08xl5 


378  x  920 


. 

°r  1S  200° 


The  small  value  both  of  exciting  current  as  compared  with  the 
full  load  one.  and  the  small  leakage  field,  show  that  the  design  is 
very  good  as  far  as  the  magnetic  properties  are  concerned. 

With  the  full  load  primary  current  of  15  amperes  the  equiva- 
lent resistance  drop  is  evidently  33'3  volts  and  the  reactance 
pressure  is  41  '7.  By  applying  the  method  given  on  p.  71  it  will 
be  found  that,  assuming  the  open-circuit  pressure  on  the  secondary 
is  100  volts,  the  pressure  with  a  full  non-inductive  load  will  be 
98'3  volts.  If  the  same  load  be  taken  but  with  a  power  factor  of 
0'6,  the  current  will  have  to  be  greater  in  the  ratio  of  5  to  3,  hence 
the  two  pressures  now  become  55  '5  and  69'5.  On  again  applying 
that  method  it  will  be  found  that  the  terminal  pressure  is  about 
96  volts. 


CHAPTER  VI. 


SPECIAL   FOEMS   OF   TRANSFORMER. 


Auto-transformer.  In  some  cases  it  is  not  necessary  to 
have  two  distinct  coils  provided  on  the  iron  core  for  the  purpose 
of  transformation.  Let  a  transformer  have  the  ratio  p  as  in 
Fig.  63,  where  p  is  3.  As  we  have  seen,  when  a  pressure  S0  is 
applied  to  the  terminals  of  the  primary  coil  p,  the  pressure  in  the 
secondary  is  SQjp.  Let  this  secondary  deliver  a  current,  then 


Fig.  63. 

very  approximately  we  know  that  the  current  in  the  primary 
being  <$,  that  in  the  secondary  will  be  p^.  Instead  of  providing 
a  second  coil  to  act  as  secondary,  let  wires  be  brought  out  from 
the  winding  of  the  coil  on  the  core  at  such  points  that  there 
is  between  them  the  same  number  of  turns  as  in  the  previous 
secondary,  as  shown  in  the  lower  figure,  such  an  arrangement 
is  called  an  auto-transformer  or  "  compensator."  Since  the  flux  is 
common  to  the  two  circuits  thus  formed,  the  ratio  of  transforma- 
tion will  be  the  same  as  before,  and  thus  as  far  as  the  transference 
of  power  between  the  two  circuits  is  concerned,  the  conditions  are 


SPECIAL    FORMS   OF   TRANSFORMER  79 

unaltered.  The  current  in  the  whole  winding  being  ^,  that  in  the 
circuit  attached  to  the  part  will  be  p^.  Since  these  currents  are, 
as  we  have  seen,  antiphased,  the  nett  current  in  the  common  turns 
will  be  (p  —  1)  ^.  Let  us  assume  that  the  current  density  in  the 
two  coils  of  the  original  transformer  was  the  same,  then  if  R 
denote  the  resistance  of  the  primary,  that  of  the  secondary  will  be 
R/p,  and  since  the  currents  are  respectively  ^  and  p%?,  the  drops 
will  be  the  same  as  far  as  ohmic  resistance  is  concerned,  and  the 
total  energy  lost  will  be  R^2(p  +  1);  as  the  auto-  transformer  has 
the  same  current  density,  the  portion  of  the  coil  which  carries  the 
primary  current  only  must  have  the  same  cross  section  as  before, 
but  its  length  will  be  less  by  the  part  carrying  the  two  currents,  that 

is,  will  be  f—     -  j  times  its  former  length  and  hence  f  —   -  ]  times 

its  former  resistance.  Hence  if  R  still  denotes  the  original  primary 
resistance,  the  resistance  of  this  part  of  the  auto-  transformer  will 

be  (<-—  -  }  R  and  with  the  same  current  ffi  as  before  the  energy 

loss  will  be  ^2R(—  -  J  .  The  common  part  of  the  winding  will 
now  carry  the  current  (p  -  1)^;  in  the  original  case  the  resistance 


p 
of  the  secondary  was  —  and  it  carried  the  current  p'ff,  hence  with 

the  same  current  density,  the  resistance  of  the  common  part  can 
now  be 


R      p  R 

or 


i    ^  i  • 

p    p-l         p-l 

Thus  the  energy  lost  in  that  part  is 

(p-1)2^2.^  or 
or  the  total  loss  will  be 


-        or 


P 

The  ratio  of  this  to  the  former  is 

p2  -  1  p-l 

-    or  "      -. 

PG>  +  1)  P 

We  thus  see  that  the  ohmic  loss  in  an  auto-transformer  is  less 
than  that  in  a  transformer  with  the  same  current  density,  and 
hence  it  can  be  made  smaller  than  the  corresponding  transformer. 
In  fact  it  will  readily  be  seen  from  the  expression  just  derived, 
that  with  a  2  to  1  ratio  the  auto-transformer  need  be  only  half  as 
large  as  the  corresponding  transformer,  while  with  the  ratio  of  3 
to  1  it  will  be  Jrds  of  the  size.  In  the  above  the  magnetising 
current  has  been  left  out  of  account. 


80  ALTERNATING    CURRENTS 

A  convenient  form  of  this  apparatus  for  many  purposes  is 
obtained  if  various  points  of  the  winding  are  brought  out  to 
form  several  secondaries  with  different  definite  pressures.  Thus, 
for  example,  if  tappings  to  the  coil  are  provided  at  ten  equal 
intervals  along  the  coil,  the  pressure  between  each  successive 
tapping  will  be  one  tenth  of  the  applied  pressure,  and  thus  the 
whole  forms  an  alternate  current  potential  divider.  In  such 
a  case  the  coil  may  for  convenience,  as  for  example  for  laboratory 
purposes,  be  wound  with  the  same  gauge  of  wire  throughout, 
or  the  gauge  may  be  varied  to  suit  the  currents  that  will  be 
required  for  the  successive  connections  to  the  tappings. 

It  is  evident  that  such  an  apparatus  is  chiefly  of  use  when  the 
ratio  of  transformation  is  small.  With  large  ratios  the  expression 
given  shows  that  little  advantage  accrues,  and  further  the  primary 
and  secondary  coils  being  perforce  connected,  there  is  danger 
of  a  high  primary  pressure  being  applied  to  apparatus  connected 
to  the  secondary  should  any  breakdown  of  insulation  occur. 

Current  transformer.  Transformers  are  sometimes  used 
in  connection  with  ammeters  of  small  maximum  range  to  enable 
large  currents  to  be  measured.  In  the  consideration  of  the  trans- 
former it  will  be  recollected  that  under  all  circumstances,  what- 
ever be  the  resistances  of  the  coils,  the  primary  current  was 
such  as  to  equilibrate  the  secondary  one,  and  provide  a  small 
part  over  to  allow  for  the  establishment  of  the  flux  in  the  iron 
core.  If  this  latter  part  be  small  enough  to  be  negligible,  then 
under  all  circumstances  the  two  currents  will  be  exactly  in  the 
ratio  of  the  turns  of  the  two  coils.  The  condition,  therefore, 
that  such  a  transformer  must  fulfil  is  simply  that  the  mag- 
netising current  is  reduced  to  the  smallest  possible  value. 
In  order  that  this  may  be  the  case,  the  iron  circuit  must  be  so 
arranged  that  the  maximum  induction  in  the  cycle,  and  con- 
sequently the  hysteretic  loss,  is  as  small  as  possible.  This  can  be 
secured  by  having  very  small  pressures  produced,  and  hence  the 
ammeter  should  be  of  as  low  resistance  as  possible  in  order  to 
require  a  small  pressure  at  its  terminals  with  the  full  reading. 
Again,  the  iron  core  must  be  designed  to  have  the  smallest 
possible  loss  with  this  applied  pressure. 

With  fair  precautions  in  design  it  is  possible  to  make  current 
transformers  in  which  the  angle  between  the  two  currents  differs 
from  antiphase  by  less  than  J  of  a  degree,  thus  ensuring  the 
exactness  of  the  ratio  of  transformation. 

Air  core  transformer.  In  some  cases  it  is  useful  to  be 
able  to  produce  a  current  that  is  practically  at  right  angles  to 
a  given  current,  and  this  can  be  secured  as  follows.  Consider 
a  transformer  in  which  there  is  no  iron  core,  then  as  we  can  readily 
see,  owing  to  there  being  no  hysteresis,  the  flux  is  accurately 


SPECIAL   FORMS   OF   TRANSFORMER 


81 


in  phase  with  this  current  when  the  secondary  circuit  is  unloaded. 
It  follows  that  the  induced  pressures,  both  primary  and  secondary, 
will  then  be  exactly  in  quadrature  with  the  current  flowing.  If  the 
secondary  be  allowed  to  carry  a  current,  this  is  no  longer  the  case, 
since  the  primary  magnetomotive  force  will  have  to  equilibrate 
this  new  one  as  well  as  produce  the  flux  between  the  coils.  Let 
the  two  coils  however  be  closely  entwined  so  that  the  flux  is 
practically  the  same  for  each,  and  let  the  secondary  be  very  lightly 
loaded,  and  it  will  be  evident  that  the  magnetomotive  force 
required  for  the  secondary  will  be  very  small,  and  hence  the 
E.M.F.  in  the  secondary  will  be  practically  in  quadrature  with  the 
primary  current;  hence  provided  the  secondary  circuit  be  non- 
inductive,  the  resulting  small  current  will  also  be  practically  in 
quadrature  with  the  main  one.  The  transformer  in  fact  is  one 
with  a  relatively  enormous  magnetising  current  and  very  small 
load  one.  In  both  this  case,  and  also  in  the  last,  when  it  is 
desired  to  cut  the  transformer  out  of  action  the  primary  should 
be  short-circuited  as  will  be  evident  from  the  consideration  of 
its  relation  to  the  circuit. 

Application  to  wattmeter.  This  property  of  an  air  cored 
transformer  has  been  used  by  I)r  Sumpner  to  enable  alternate 
current  wattmeters  to  be  made  with  iron  cores.  Consider  the 
case  of  Fig.  64.  D  is  a  D'Arsonval  galvanometer  but  made 


Fig.  64. 

with  laminated  field  magnets  and  as  small  an  air  gap  as  possible. 
The  winding  on  it  is  made  of  wire  of  as  large  a  guage  as  can 
conveniently  be  used  for  the  instrument,  so  as  to  keep  the  ohmic 
resistance  of  the  winding  as  small  as  possible,  and  this  winding  is 
put  across  the  mains  supplying  the  power.  T  is  a  small  air  core 
transformer  of  the  type  we  have  just  considered,  having  a  fairly 

L.  6 


82  ALTERNATING   CURRENTS 

large  ratio  of  transformation.  The  primary  of  this  transformer  is 
put  in  one  of  the  mains  supplying  current  to  the  load,  the 
secondary  being  in  series  with  the  coil  of  the  D'Arsonval  and 
a  high  non-inductive  resistance  R. 

Owing  to  the  pressure  applied  to  the  ends  of  D  an  alternating 
current  will  flow  in  its  winding  which  will  generate  an  alternating 
flux  in  the  whole  magnetic  circuit,  and  this  again  will  necessitate 
the  production  of  an  induced  E.M.F.  exactly  in  quadrature  with 
the  flux ;  if  the  resistance  of  the  winding  on  D  be  very  low,  so 
that  the  ohmic  drop  due  to  the  magnetising  current  is  negligible, 
the  applied  pressure  being  in  antiphase  with  this  induced  pressure 
will  likewise  be  in  quadrature  with  the  flux.  Let  B  be  the  in- 
stantaneous value  of  the  intensity  of  the  induction  in  the  air  gap, 
the  total  flux  will  be  proportional  to  B,  further  with  the  usual 
arrangement  of  a  uniform  air  gap  the  value  of  B,  the  intensity  of 
flux  in  which  the  moving  coil  is  situated,  will  be  independent  of 
the  angular  position  of  that  coil,  but  solely  dependent  of  the  total 
flux  round  the  circuit.  Hence  if  e  is  the  instantaneous  applied 

,          .,  dB        ,    . 

pressure  we  can  very  approximately  write  e  =  a.  -^— ,  a  being  a 

constant  depending  on  the  magnetic  circuit  only.  If  c  denote  the 
instantaneous  current  in  the  primary  of  the  transformer,  T,  the  flux 
will  be  proportional  to  that  current  provided  the  current  taken  by 
the  secondary  attached  to  the  coil  of  D  is  very  small,  which  will 
be  the  case  if  R  is  large.  Hence  the  E.M.F.  induced  in  the 

cLc 

secondary  will  be  proportional  to  -^ .  Thus  if  the  secondary  is  en- 
tirely non-inductive  the  current  in  the  coil  will  be  given  by 
d  =  -ft .  -r.  where  A;  is  a  constant.  Hence  the  torque  on  the  coil 

being  proportional  to  the  mean  value  of  the  product  of  the 
induction  in  the  gap  into  the  current  in  the  coil,  will  be  pro- 
portional to  the  expression 


This  can  be  written 

k    1  f'       dc 


or,  by  using  the  theorem  on  mean  values  given  on  p.  33,  reduces 
to  the  form 

k    1  f      dB    ,. 

-n  .  -          C   .    -j-.dt. 

R  r  J  o       dt 
Which  immediately  leads  to 

k    1  fT 

— n •  -  I    c .e  .at, 
aR  rj  o 


SPECIAL   FORMS   OF  TRANSFORMER  83 

k 
or  -~  .  Wy  where  W  is  the  mean  power,  hence  the  torque  will  be 

proportional  to  that  mean  power.  With  a  control  of  the  ordinary 
type  it  will  follow,  as  in  the  ordinary  D'  Arson  val,  that  the  power 
taken  by  the  load  will  be  proportional  to  the  deflection. 

Ammeter.  It  can  be  seen  that  the  same  type  of  instrument 
can  be  used  as  an  ammeter.  Let  D  be  wound  with  a  few  turns 
and  placed  in  series  in  the  circuit,  in  which  is  also  placed  a  low 
resistance  on  which  the  coil  of  the  instrument  is  connected  as 
a  shunt.  If  this  shunt  circuit  be  practically  non-inductive,  which 
can  readily  be  arranged  by  means  of  a  series  resistance,  the  current 
in  the  coil  will  be  nearly  in  phase  with  the  current  in  the  main. 
The  flux  in  D  will  be  proportional  to  the  current  for  all  ordinary 
values  of  the  latter  since  the  reluctance  is  principally  due  to  air. 
Further  the  angle  of  hysteretic  advance  will  be  practically  constant 
as  well.  Hence  both  the  induction  in  the  gap  and  the  current  in  the 
coil  will  be  proportional  to  the  main  current  while  the  phase  angle 
between  them  is  nearly  constant.  It  follows  that  the  torque  is 
practically  proportional  to  the  square  of  the  current  exactly  as  in 
the  ordinary  dynamometer. 

Voltmeter.  In  order  to  use  the  instrument  as  a  voltmeter 
the  winding  of  D  is  placed  in  shunt  on  the  mains  as  for  the  watt- 
meter, but  the  coil  is  now  put  in  series  with  a  condenser,  the  two 
being  also  put  across  the  mains.  As  before  the  relation  between 
the  air  gap  induction  and  the  applied  pressure  will  still  be  given  by 

dB 


when  the  resistance  of  the  winding  is  very  small.  Let  F  be  the 
capacity  of  the  condenser,  and  assume  that  it  is  not  quite  perfectly 
insulating  so  that  it  may  also  be  considered  as  possessing  a  fairly 
high  resistance  R.  The  current  in  the  condenser  circuit  and 
hence  in  the  coil  will  be  given  by 


thus  the  torque  experienced  by  it  will  be  proportional  to 

-['B.c.dt, 


that  is  to  ,   ^  ,  JL 

at 

"Consider  the  last  integral,  it  is  evidently  equal  to 

dB 


6—2 


84  ALTERNATING    CURRENTS 

and  thus  from  p.  33  is  zero.     The  integral  thus  reduces  to 

F  TR  de 
TJo       d*       ' 

or  from  the  same  page  we  see  that  it  is  equal  to 

-  I  e.-j-.dt. 

r  JQ       dt 

On  substituting  for  -^-  this  leads  to 

F  i  r 

a'rJo6 

and  hence  the  torque  is  proportional  to  the  square  of  the  virtual 
pressure.  With  a  control  of  the  same  form  as  before  it  follows 
that  the  deflection  will  be  a  measure  of  this  square. 

It  will  be  noted  that  no  assumption  as  to  the  form  of  any  of 
the  quantities  has  been  made,  and  hence,  within  the  limits  of  error 
of  the  apparatus,  it  is  suitable  for  measuring  the  different  quantities 
for  any  form  of  curve. 

It  can  also  readily  be  seen  that  the  relative  calibration,  that  is, 
the  connection  between  the  product  of  the  flux  and  the  coil  current, 
can  be  made  with  direct  currents.  All  that  is  necessary  is  to  excite 
the  magnet  with  a  fixed  current  and  then  send  known  small  direct 
currents  through  the  coil.  If  the  deflections  corresponding  to  these 
currents  be  noted,  this  will  evidently  constitute  the  relative  cali- 
bration of  the  instrument.  The  actual  constant  can  then  be 
adjusted  for  any  desired  power,  current  or  pressure  by  proper 
adjustment  of  the  resistances  or  capacities  of  the  coil's  circuit. 


CHAPTER  VII. 

LOSSES   IN  TRANSFORMERS. 

Losses  and  efficiency.  The  losses  incident  to  the  operation 
of  a  transformer  are  the  same  in  character  as  those  found  in  the 
operation  of  the  ordinary  direct  current  apparatus.  We  have  first 
the  ohmic  losses  due  to  the  currents  in  the  two  coils,  and  secondly 
the  loss  of  energy  due  to  the  cyclic  changes  of  magnetism  in  the 
core.  Since  the  condition  of  operation  is  that  the  terminal 
pressure  is  constant  and  its  periodicity  is  constant,  and  since  we 
have  seen  that  in  an  actual  transformer  the  terminal  pressure  is 
very  nearly  equal  to  the  primary  induced  E.M.F.,  it  must  follow  that 
since  the  latter  E.M.F.  is  equal  to  the  rate  of  change  of  flux  in  the 
primary  coil,  the  flux  must  alternate  with  the  same  maximum 
value  B  for  all  loads,  provided  only  the  pressure  and  its  periodicity 
be  kept  constant.  But  the  hysteretic  losses  will  be  proportional 
to  the  periods  n  and  will  be  nearly  as  the  l'6th  power  of  the 
induction  ;  further  any  eddy  current  loss  that  exists  will  be  pro- 
portional to  the  square  of  the  induction  and  of  the  periods. 

Hence  if  v  is  the  volume  of  the  core  in  cubic  centimetres  and 
A  is  a  constant  giving  the  hysteretic  loss  per  cubic  centimetre  at 
unit  periodicity,  k  being  a  similar  constant  for  the  eddy  currents, 
we  can  write  the  power  lost  in  the  iron  core  in  the  form 

WL  =  h.v.nBl*  +  k.v.n*B*, 
or  for  a  definite  core 


Hence  in  the  case  where  both  periodicity  and  pressure  are  constant 
the  total  loss  of  energy  in  the  core  will  be  very  nearly  constant, 
and  practically  independent  of  the  load  on  the  transformer. 

The  ohmic  loss  will  be  dependent  solely  on  the  current  that  is 
being  taken,  and  will  be  readily  found  from  a  knowledge  of  the 
core's  resistance.  At  any  but  the  smallest  loads,  it  is  practically 
proportional  to  the  square  of  the  secondary  current. 

Maximum  efficiency.  The  maximum  efficiency  occurs  when 
the  core  loss  and  ohmic  loss  are  the  same.  Let  WL  be  the  constant 
core  loss,  and  let  the  total  ohmic  loss  be  a.ffi,  where  a  is  a  constant 


86  ALTERNATING  CURRENTS 

and  ^P  the  secondary  current.  Let  the  secondary  pressure  be  £ 
and  the  power  factor  given  by  cos  X.  The  output  is  then  £*$  cos  X 
and  the  input  is 


The  efficiency  is  given  by 
rj  = 


Hence  the  maximum  efficiency  is  given  by  -^  =  0,  leading  to 


'  cos  X  +  WL  +  a .  ffi2)  £  cos  X  =  £9%  cos  X  (S  cos  X  +  2a .  *$), 

Tests  of  efficiency.  As  with  direct  current  machines  we 
have  three  methods  of  test  possible,  firstly  the  method  of  measuring 
the  input  and  output,  secondly  of  merely  measuring  the  losses  in 
the  apparatus  under  the  given  supply  conditions,  thirdly  of  coupling 
two  similar  pieces  of  apparatus  together  and  circulating  power 
between  them,  at  the  same  time  measuring  the  loss  of  energy. 

Direct  measurement  of  efficiency.  The  first  method  can 
be  very  shortly  dealt  with.  All  that  has  to  be  done  is  to  measure 
the  input  and  output  with  wattmeters.  If  the  secondary  load  be 
non-inductive,  a  voltmeter  and  an  ammeter  can  be  used  in  place 
of  a  wattmeter,  but  one  must  still  be  used  for  the  primary  circuit. 

Stray  power  method,  one  transformer.  The  second,  or 
stray  power  method,  consists  in  determining  the  losses  separately 
and  deducing  the  efficiency.  This  method  has  been  incidentally 
referred  to  in  Chapter  V,  but  for  completeness  will  be  again 
described.  For  simplicity  we  will  consider  that  one  coil  is  intended 
to  have  a  pressure  of  1000  volts,  the  other  a  pressure  of  100  volts. 
It  is  evident  that  whether  we  apply  1000  volts  to  the  one  coil  or 
100  volts  to  the  other  the  core  will  be  subjected  to  the  same 
cycle  of  magnetism,  the  periodicity  being  the  same  in  the  two 
cases.  Hence  the  core  loss  can  be  found  by  the  open  circuit  test 
as  follows.  Connect  the  100  volt  coil  to  mains  at  that  pressure 
and  measure  the  energy  taken.  This  will  include  two  losses,  the 
core  loss  itself  and  an  ohmic  loss  incident  to  the  passage  of  the 
no  load  current.  Since  the  loss  of  energy  in  ohmic  resistance 
is  proportional  to  the  square  of  the  current,  and  since  we  have  seen 
that  the  magnetising  current  is  very  small  compared  with  the  full 
load  current,  the  ohmic  loss  due  to  the  passage  of  the  magnetising 
current  is  negligibly  small.  Hence  this  measurement  can  be 
taken  as  giving  the  constant  core  loss  of  the  transformer,  on 
whichever  coil  it  be  working  with  as  primary.  It  remains  to  find 
the  ohmic  loss.  This  is  done  by  means  of  the  short  circuit  test. 
Connect  the  terminals  of  the  100  volt  coil  directly  by  a  wire.  Then 


LOSSES   IN   TRANSFORMERS  87 

join  up  the  1000  volt  coil  to  the  mains  at  100  volts,  connecting 
in  a  wattmeter,  a  suitable  ammeter,  and  a  series  adjustable 
resistance.  The  E.M.F.  induced  in  the  secondary  will  now  be 
concerned  solely  in  forcing  the  current  against  the  impedance 
of  the  secondary  coil  and  this  will  require  a  very  small  pressure. 
Let  the  resistance  in  the  primary  be  so  adjusted  as  to  permit  any 
required  current  to  flow  in  it,  say  the  full  load  current,  and  measure 
by  the  wattmeter  the  power  taken.  This  will,  as  before,  include 
two  parts,  the  ohmic  losses  in  both  coils  and  a  certain  core  loss. 
The  latter  will  be  dependent  on  the  flux  in  the  core  and  will  vary 
as  the  1*6 th  power  of  that  flux.  But  the  flux  is  nearly  propor- 
tional to  the  pressure  applied,  and  we  saw  that  instead  of  the 
normal  pressure  of  1000  volts  only  a  very  small  pressure  will  be 
required  on  the  coil,  hence  the  core  loss  in  this  case  will  be 
negligible  in  the  same  way  as  the  ohmic  loss  was  in  the  first 
case.  We  have  thus  sufficient  data  to  determine  for  any  assumed 
load  the  core  loss  and  the  various  total  ohmic  losses.  The  efficiency 
curve  can  then  be  found  as  follows.  Assume  a  constant  pressure 
at  the  secondary7  terminals,  say  100  volts.  The  output  or  non- 
inductive  load  will  be  the  product  of  any  current  taken  into  the 
assumed  pressure  of  100  volts.  To  each  output  add  the  constant 
core  loss  and  the  appropriate  total  ohmic  loss,  and  the  sum  will 
be  the  input.  The  ratio  of  the  two  will  give  the  efficiency.  The 
results  of  such  a  test  deduced  from  the  data  on  p.  76  are  given  in 
the  table  below. 

Constant  loss  =  400  watts. 

Copper  loss  at  full  load  =  500  watts. 

Secondary  Secondary  C-R  Total  Input  Efficiency 

current  power          loss  watts    loss  watts  °/0 

300  amp.  30    kw.  500  900  30'9    kw.  97*0 

225     „  22|    „  272  672  23'17    „  97'2 

150     „  15      „  125  525  15'52    „  96'6 

112     „  11|    „  31  431  11-68    „  96-4 

30     „  3      „  5  405  3-41    „  88'0 

This  method  has  the  advantage  that  it  can  be  used  when  only 
a  single  transformer  is  available,  but  it  does  not  at  all  test  the 
apparatus  under  the  normal  working  conditions  since  the  coils 
never  have  both  the  full  current  flowing  and  the  full  pressure 
acting.  This  is  secured  in  the  next  method,  which  however 
necessitates  the  provision  of  two  identical  transformers. 

Combined  test,  two  transformers.  The  third,  or  combined 
test,  can  be  best  explained  by  the  following  preliminary  method. 
Let  two  similar  transformers,  I  and  II  (Fig.  65),  be  taken,  but  let 
one  of  them  have  one  coil  provided  with  extra  terminals  so  that 
other  pressures  than  the  normal  one  are  available.  For  example,  let 
the  normal  pressure  on  one  of  the  coils  be  100  volts  but  let  wires 


88 


ALTERNATING   CURRENTS 


be  brought  out  at  such  different  points  as  to  give  in  addition 
pressures  falling  by  1  volt  to  say  93  volts.  Let  these  coils  be  put 
in  parallel  mains  with  a  wattmeter  connected  as  shown.  Then 


Fig.  65. 

if  the  other  pair  of  coils  be  connected  in  series  with  their  pressures 
opposing,  no  current  will  flow  when  the  full  number  of  turns  are 
employed  in  II  and  the  wattmeter  will  read  the  loss  of  energy  in 
the  two  cores.  Now  let  the  connections  be  so  made  that  the 
transformer  with  the  different  terminals  produces  only  99  volts 
while  the  other,  of  course,  still  produces  100.  The  one  volt 
difference  will  then  be  available  to  circulate  power  between  the 
two  primaries  and  hence  between  the  secondaries.  The  wattmeter 
will  then  read  the  loss  of  energy  incident  to  the  transformation. 
The  current  circulating  will  be  given  by  the  ammeter  joined 
in  the  secondary,  and  the  losses  found  with  that  current  flowing. 
By  proceeding  in  this  way  we  can  find  the  power  taken  up  to  full 
load  current,  provided  sufficient  terminals  are  available  on  the 
transformer.  A  correction  may  be  made  for  the  small  loss  in  the 
ammeter. 

This  method  is  manifestly  useless  in  the  case  where  the  two 
transformers  are  of  the  ordinary  commercial  type  without  the 
special  terminals  on  one  of  them.  In  such  a  case  some  other 
method  must  be  employed  to  circulate  the  power  between  them. 
For  this  purpose  a  small  auxiliary  transformer  is  used  connected 
to  the  circuit  as  shown  in  Fig.  66.  I  and  II  are  the  two  trans- 


LOSSES   IN   TRANSFORMERS 


89 


formers  under  test  having  their  high  pressure  coils  joined  in 
parallel.  An  auxiliary  transformer  0  has  its  primary  connected 
across  the  mains  in  series  with  an  adjustable  resistance  R,  while 
its  secondary  is  placed  in  series  with  the  other  coil  of  one  of  the 
transformers,  I;  the  transformer  II  has  its  free  coil  placed  in 
parallel  with  the  other  two  on  the  supply  mains.  Two  wattmeters 
are  employed,  the  one  W^  being  so  joined  that  the  power  taken  by 
the  transformer  C  is  not  measured  by  it,  the  other  Wz  being 


connected  so  as  to  measure  the  power  delivered  by  the  secondary 
of  the  auxiliary  transformer,  C.  It  will  be  seen  that  by  altering 
the  value  of  R  different  pressures  can  be  supplied  to  the  primary 
of  G  and  thus  any  desired  pressure  produced  in  its  secondary. 
Thus  the  pressure  in  the  primary  circuit  of  I  can  be  made  to  differ 
by  any  desired  amount  from  the  pressure  in  the  primary  circuit  of 
II,  and  hence  a  current  of  any  desired  value  can  be  made  to 
circulate  in  transformers,  and  this  can  be  measured  by  the  ammeter 
A.  The  pressure  at  which  this  current  power  is  supplied  is  given 
by  the  voltmeter  V.  From  the  method  of  connection  it  is 
seen  that  Wl  will  measure  the  core  loss  in  the  two  transformers, 
while  Wz  will  measure  the  power  that  has  to  be  supplied  to 
circulate  the  power  between  the  transformers  A  and  B,  or  W 2  will 
measure  the  ohmic  loss.  The  loss  in  the  ammeter  may  be  allowed 


90  ALTERNATING    CURRENTS 

for  as  in  the  last  cases,  and  hence  the  total  nett  loss  determined  for 
any  current  in  the  transformers.  The  loss  can  then  be  allocated 
one-half  to  each  and  the  efficiency  readily  deduced  in  the  ordinary 
way. 

Rejection  of  lost  energy.  All  the  energy  lost  in  a  trans- 
former is  necessarily  rejected  as  heat,  and  in  small  transformers 
this  rejection  is  made  by  the  ordinary  processes  of  radiation. 
For  the  sake  of  safety  the  apparatus  is  generally  contained  in 
an  iron  case  which  is  often  ribbed  to  facilitate  the  radiation.  The 
actual  transference  of  the  heat  from  the  transformer  to  the  case  is, 
under  these  •  circumstances,  brought  about  by  convection  currents. 
in  the  air  in  the  -case,  and  to  provide  a  better  medium  for  this 
purpose  the  transformer's  case  is  often  filled  with  oil.  This  has. 
the  additional  advantage  of  maintaining  the  insulating  properties 
of  the  covering  to  the  wires,  and  carrying  the  heat  direct  from  the 
metal  surface  instead  of  from  that  of  the  insulation.  The  rise  of 
temperature  will  depend  on  the  load  carried,  and  since  the  rate  of 
lass  of  heat  due  to  radiation  is  practically  proportional  to  the 
temperature  rise,  while  the  losses  are  constant  as  far  as  the  core 
loss,  and  proportional  to  the  square  of  the  current  for  the  ohmic 
loss,  this  rise  of  temperature  will  increase  more  rapidly  than  the 
load.  It  is  found  in  practice  that  if  it  exceeds  a  superior  limit  in 
the  neighbourhood  of  70°  C.,  progressive  deterioration  takes  place 
in  the  insulation.  The  test  of  a  transformer  should,  then,  include 
a  measure  of  its  temperature  rise  after  definite  conditions  of  load 
have  been  maintained  for  definite  periods.  This  can  be  done  by 
thermometers  placed  in  contact  with  the  parts  whose  temperature 
is  required  to  be  known  but,  in  the  case  of  the  windings,  is  best 
found  from  a  measurement  of  their  resistance  in  the  usual  way. 
From  the  known  value  of  the  coefficient  of  increase  of  resistance 
of  copper,  that  is  0*4  per  cent,  per  degree  Centigrade,  the  required 
temperature  can  be  at  once  determined:  this  method  has  the 
advantage  of  giving  the  actual  temperature  of  the  copper,  which 
must  be  somewhat  in  excess  of  the  surface  temperature  of  the 
cores,  owing  to  the  necessary  existence  of  a  temperature  gradient 
between  the  inside  and  outside  of  the  coils. 

With  large  transformers  additional  precautions  must  be  taken 
to  ensure  a  safe  temperature.  For  let  two  transformers  be  taken 
working  at  the  same  induction  and  the  same  current  density,  then 
the  average  loss  will  be  the  same  in  each  per  cubic  centimetre, 
but  the  area  available  for  rejecting  the  lost  energy  only  increases 
as  the  square  of  the  dimensions ;  hence  in  such  a  case  the  trans- 
former of  larger  size,  when  loaded  in  the  same  proportion  as  the 
smaller,  must  exceed  it  in  temperature.  This  is  partly  avoided  by 
using  a  somewhat  lower  induction  and  current  density  in  the 
larger,  but  in  addition  special  means  are  adopted  to  reject  the 
heat.  This  is  done  by  forced  circulation  of  air  through  the  case, 


LOSSES   IN   TRANSFORMERS  91 

or  when  oil  filled  cases  are  used,  a  system  of  pipes  is  placed  in 
them  through  which  cold  water  is  circulated. 

Effect  of  temperature  on  the  core  loss.  The  effect  of  an 
increase  of  temperature  is  to  increase  the  ohmic  resistance  of  the 
eddy  current  circuits  and  hence  will  tend  to  diminish  the  loss  of 
energy  in  the  core  that  results  from  them.  The  effect  on  the 
hysteresis  loss  is  in  some  cases  a  secular  one,  as  it  is  known  that 
certain  descriptions  of  iron  show  a  gradual  and  considerable 
increase  of  hysteretic  loss  when  subjected  to  a  fair  temperature 
for  some  time.  Such  an  effect  can  only  be  detected  by  tests  of  the 
core  loss  conducted  when  the  transformer  is  made,  and  after  a 
considerable  period  of  working  has  elapsed.  Modern  improvements 
in  the  manufacture  of  iron  have  largely  diminished  this  source  of 
increased  loss. 

Loss  in  iron,  wattmeter  method.  The  no  load  test  of 
a  transformer  or  choking  coil  gives,  as  we  have  seen,  the  core  loss 
of  the  same  ;  if  we  wish  to  find  the  loss  in  a  given  sample  of  iron, 
we  have  merely  to  make  it  into  such  a  choking  coil  and  use  the 
wattmeter  method  of  measuring  the  loss  of  energy  in  a  given 
sample  of  iron,  one  which  has  many  advantages.  The  sample  of 
iron  having  been  made  up  into  a  choking  coil,  is  connected  to  a 
source  of  current  giving  as  nearly  as  possible  a  sine  curve  of 
pressure  at  a  known  periodicity  n.  In  the  circuit  is  placed  a 
wattmeter.  The  choking  coil  is  provided  with  a  secondary  circuit 
of  fine  wire  of  a  suitable  number  of  turns  T  to  which  is  attached 
an  electrostatic  voltmeter.  Various  pressures  are  applied  to  the 
primary  terminals  and  in  each  case  the  power  taken  and  the 
pressure  are  read.  The  power  will  be  that  absorbed  by  the  iron 
core  together  with  the  ohmic  loss  in  the  circuit.  The  latter  can 
be  allowed  for  if  an  ammeter  be  placed  in  series  so  that  the  loss 
in  the  winding  can  be  calculated  from  its  measured  resistance.  The 
nett  loss  in  watts  divided  by  the  volume  in  cubic  centimetres  and 
the  periods  per  second  gives  the  ergs  per  c.c.  per  cycle  lost  in  the 
core.  The  induction  produced  can  be  found  as  follows  :  if  £  be  the 
reading  of  the  voltmeter,  since  the  pressure  produced  will  be  very 
nearly  the  same  in  form  as  the  applied  pressure,  that  is  sinusoidal, 
it  will  follow  that  the  maximum  pressure  will  be  *J%£.  But  if  B  is 
the  maximum  induction  in  the  iron  core,  the  section  being  S  we 
know  that  the  maximum  pressure  will  be  B  .  S  .  T  .  ZTTH,  hence  we 
have 

£  x  108 


The  area  of  the  iron's  section  is  best  found  by  determining  the 
specific  gravity  of  the  sample  and  the  linear  dimensions  of  the 
core  ;  if  the  weight  of  the  whole  be  then  found  the  section  can 


92 


ALTERNATING   CURRENTS 


be  at  once  calculated.  Hence  we  can  find  the  maximum  induction 
in  the  iron  and  thus  the  relation  between  this  maximum  induction 
and  the  nett  loss  per  cubic  centimetre  per  cycle.  The  ordinary 
circular  washers  offer  some  difficulty  in  winding  for  different  tests 
and  it  is  desirable  to  avoid  this  if  possible.  The  student  will  find 
described  in  a  paper  by  Mr  G.  F.  C.  Searle,  Journal  I.  E.  E.  vol.  34, 
a  form  of  magnetic  circuit  which  enables  strips  to  be  used  and  does 
not  necessitate  rewinding  each  sample  for  test. 

Example.  The  curve  given  in  Fig.  67  was  obtained  in  this  way, 
and  as  an  example  of  the  method  of  reduction  the  following  details 
may  be  given. 


sooo 


6000 


2000 


2000 


4000          6000 

M&ximum  Value  of  B. 


sooo 


10000 


Fig.  67. 

Section  of  iron  =  21'3  sq.  cm. 

Volume  of  iron  =  1563  c.c. 

Resistance  of  magnetising  coil,  R  =  0115  ohms. 

Turns  in  the  secondary  coil  =  200. 

Periods  per  second  =  85. 

If  <£"  is   the   reading   of  the   voltmeter  the  constant  giving 
B  from  its  readings  is  then 

108 
V2x,rx  85x200x21-3  *  °r  62'5  * 


LOSSES   IN   TRANSFORMERS  93 

If  ^  is  the  current  and  W  the  watts  in  the  same  case,  the  nett 
watts  will  be  Wn  —W  —  ffiR,  and  the  ergs  per  cubic  centimetre 
per  cycle  will  be 

Wn  x  107 


85  x  1563 


or 


In  one  case  the  pressure  was  74  volts,  the  total  power  23*8 
watts,  and  the  current  1*6  amperes,  giving  23'5  for  Wn.  Hence 
the  induction  is  about  4630  and  the  loss  per  c.c.  per  cycle  is  1766 
ergs.  A  set  of  similar  observations  were  taken  from  which  the 
numbers  plotted  were  found. 


CHAPTER   VIII. 

THE   SERIES   MOTOR. 

WHEN  the  consideration  of  the  properties  of  the  rotating  field 
is  undertaken  it  will  be  seen  that  by  the  use  of  the  same  it  is 
possible  to  obtain  a  satisfactory  motor,  and  that  such  a  motor  will 
possess  very  closely   the  characteristic  properties   of  the   direct 
current  shunt  motor,  that  is,  it  will  maintain  nearly  constant  speed 
up  to  its  full  load.     For  many  purposes  such  a  motor  fulfils  the 
required  conditions,  but  in  certain  cases,  such  as  for  rapid  and 
frequent  accelerations  of  tram-cars,  etc.,  the  paramount  feature  of 
the  motor  necessary  is  rather  the  production  of  a  large  torque  when 
the  speed  is  slow,  and  a  comparatively  small  one  when  the  speed  is 
high.     The  above  motor  cannot  without  certain  additions  even 
approximately  fulfil  these  conditions  without  excessive  waste  of 
energy,  and  for  many  purposes  it  would  be  desirable  to  have  a 
motor  possessing  the  valuable  properties  of  the  ordinary  series 
motor,  especially  when  traction  is  the  object.     Let  a  series  motor 
be  supplied  with  alternating  currents,  since  both  the  field  and  the 
armature  fluxes  change  with  the  current,  the  torque  will  always  be 
in  the  same  direction,  and  hence  such  a  motor  would  produce  a 
definite  positive  torque,  but  certain  alterations  must  be  made  in 
its  construction.     Firstly,  the  field  must  be  laminated  as  well  as 
the  armature  to  avoid  eddy  current  losses.    Secondly,  to  reduce  the 
self-induction   the   field   must   be   wound  with  as  few  turns  as 
possible,  and  this  would  entail  the  armature  having  more  turns. 
Thirdly,  the  armature  when  carrying  the  alternating  current  will 
produce  an  alternating  flux  of  the  same  nature  as  the  cross  flux 
produced  by  a  direct  current  armature,  its  direction  being  at  right 
angles  to  the  main  flux.     The  armature  conductors  in  cutting  this 
flux  would  evidently  produce  an  E.M.F.  which  would  have  to  be 
equilibrated  by  the  main  pressure.     This  flux  can  be  annulled  by 
a  compensating  winding  put  in  series  with  the  armature,  but  so 
connected  as  to  oppose  the  magnetic  effect  of  the  armature,  and 
having  such  a  number  of  turns  that  the  total  number  of  ampere- 
turns  in  it  is  equal  to  those  on  the  armature,  as  shown  at  W,  W  in 
Fig.  68.  In  the  present  case,  since  the  current  is  alternating,  it  is  not 


THE    SERIES   MOTOR 


95 


necessary  to  actually  put  this  coil  in  series,  it  can  consist  of  a  set 
of  short-circuited  windings,  the  current  in  them  being  induced  by 
the  varying  flux  due  to  the  armature,  so  that  this  auxiliary  winding 


Fig.  68. 

acts  with  respect  to  the  armature  as  a  short-circuited  secondary 
would  to  a  transformer  and  hence  nearly  annuls  the  armature's 
flux.  The  flux  from  the  field  will  evidently  produce  no  current 
in  this  short-circuited  coil  since  on  the  whole  it  is  not  cut  by 
that  flux,  and  thus  there  will  be  no  nett  E.M.F.  produced  in  the 
coil  by  the  field  flux.  In  this  way  it  is  possible  to  reduce  the 
self-induction  of  the  whole  motor  to  practically  that  of  the  field 
coils  only. 

Now  consider  the  action  of  the  field  flux  on  the  armature ;  if 
this  be  at  rest,  since  the  flux  passing  down  ABC  is  the  same 
as  that  passing  down  ADC,  and  since  the  conductors  in  those 
two  halves  cut  this  flux  in  the  opposite  direction,  there  is  no  nett 
E.M.F.  produced  in  the  armature  by  the  direct  action  of  the  field 


96  ALTERNATING   CURRENTS 

flux.  When  rotation  takes  place,  an  E.M.F.  will  be  produced  in 
exactly  the  same  way  as  in  the  direct  current  motor  and  this  E.M.F. 
will,  as  in  that  case,  be  proportional  to  the  speed  and  to  the  flux 
of  magnetism  due  to  the  field,  though  it  will  necessarily  be  an 
alternating  E.M.F.  Thus  if  the  total  flux  cut  by  the  armature  at 
any  moment  be  <£  and  if  the  armature  be  rotating  at  n  revolutions 
per  second  and  have  T  conductors  on  its  periphery,  the  E.M.F.  pro- 
duced, from  analogy  with  the  direct  current  case,  will  be  (f).n.T,so 
that  so  far  there  is  much  the  same  state  of  things  existing  as  in  a 
direct  current  motor.  Now  consider  the  state  of  affairs  in  a  coil 
that  is  undergoing  commutation,  which  will  take  place  in  the 
neutral  zone  since  the  armature  reaction  is  annulled.  As  in  the 
direct  current  motor  it  will  be  necessary  to  commute  the  current 
in  the  coil,  but  in  addition  it  will  be  seen  that  while  the  coil  is  in 
the  position  of  commutation  it  is  situated  in  such  a  way  relative 
to  the  field  magnet  that  it  is  experiencing  the  full  flux  of 
magnetism  from  the  same,  it  must  thus  be  acting  as  a  short- 
circuited  secondary  of  a  transformer  and  hence  very  heavy  currents 
can  be  generated  in  it ;  thus  there  will  be  an  entirely  new  factor 
to  consider  in  commutation.  This  effect  can  be  to  a  large  extent 
overcome  in  several  ways ;  one  method  is  to  connect  the  coils  to  the 
commutator  by  strips  that  have  a  higher  resistance  than  is  the  case 
in  a  direct  current  motor  (see  Fig.  68,  lower  half):  during  the 
commutation  period  this  interposes  a  high  resistance  in  the  circuit 
of  the  coils  under  the  brush  and  thus  prevents  the  currents  induced 
having  large  values.  As  regards  the  main  current  the  resistance 
added  to  the  armature  is  less  than  that  in  the  local  circuits, 
as  is  evident  from  the  figure,  where  it  will  be  seen  that  these 
strips  are  in  series  with  regard  to  the  local  circuit  formed  by  a  coil 
under  a  brush,  but  in  parallel  as  far  as  the  main  current  flowing 
up  to  the  brush  is  concerned.  Other  devices,  such  as  providing 
reactances  in  the  circuits  of  the  armature  coils  which  are  in  series 
for  the  position  when  the  coil  is  under  the  brush  but  annul  one 
another  as  far  as  the  main  current  is  concerned,  or  the  provision  of 
some  form  of  reversing  pole-piece,  have  been  used  with  success  in 
diminishing  to  a  very  great  extent  the  commutation  difficulties. 

The  following  graphical  construction  can  be  derived  under 
certain  assumptions  for  representing  the  operation  of  the  motor. 
The  total  flux  impressed  on  the  circuit  can  be  assumed  to  vary  in 
a  sine  manner  and  to  be  given  by  $  =  <I>  sin  pt.  This  flux  is  due 
to  the  current  that  the  motor  is  taking,  and  since  there  is  an  air 
gap  in  the  circuit,  the  angle  of  hysteretic  lead  will  be  small,  so 
that  we  may  very  approximately  write  the  current  as  being  given 
by  c  =  C  sin  pt.  Let  us  assume  that  the  whole  machine  can  be 
treated  as  if  it  had  a  definite  resistance,  R,  and  a  definite  reactance, 
8,  the  former-  being  such  as  to  allow  for  all  the  losses  of  energy  in 
the  same.  This  cannot  be  exactly  true,  the  ohmic  resistance  will 
be  nearly  constant,  but  the  losses  incident  to  rotation  and  to 


THE    SERIES   MOTOR 


97 


hysteresis  will  evidently  not  be  so.  If  the  field  be  non-saturated, 
as  is  practically  the  case,  the  reactance  will  be  constant  or  very 
nearly  so.  On  this  assumption  the  current  will  be  accompanied 
by  two  pressures,  the  one  CR .  sin  pt  in  phase  with  it,  the  other 
CS .  cos  pt  in  quadrature.  The  vectors  representing  these  two 
quantities  will  evidently  preserve  a  constant  angular  relation,  and 
that  between  their  resultant  and  the  vector  for  the  reactance 
pressure  will  likewise  be  constant.  This  will  be  denoted  by  <f>. 
There  is  in  addition  the  E.M.F.  due  to  the  rotation  of  the  armature, 
which  in  this  case  will  be  n .  T .  <& .  sin  pt.  Hence  we  may  show  the 
relation  between  the  quantities  as  in  Fig.  69.  The  line  0V 


B 


Fig.  69. 


represents  the  applied  pressure  of  which  the  two  components  are 
(RC  4-  nT<&)  sin  pt  and  S.C.cospt.  On  this  line  draw  a  semi- 
circle and  let  S  be  any  point  on  it,  the  lines  VS  and  OS  are  the 
two  components,  let  VS  be  S .  C,  then  OS  is  (R .  C  +  nT.  <£>).  If 
the  part  SE  be  cut  off  equal  to  R  .C,  it  follows  that  the  other  part 
OE  is  n .  T .  4>,  that  is,  OE  represents  the  E.M.F.  due  to  the  rotation 
of  the  armature.  Since  SE  is  the  vector  representing  the  pressure 
required  for  the  resistance,  it  follows  that  it  can  also  be  taken  to 
represent  the  current  to  some  appropriate  scale;  the  aagle  SVE  is 
evidently  the  constant  angle  <£  referred  to  above.  Now  draw  the 
line  VQ  making  this  same  angle,  </>,  with  0  V  that  VS  makes  with 


98 


ALTERNATING    CURRENTS 


VE,  and  draw  the  line  OB  perpendicular  to  0  V.  A  semicircle  on 
OB  will  evidently  pass  through  Q  since  BQO  and  VQO  are  both 
right  angles,  this  semicircle  will  cut  SO  in  R  and  we  will  first  show 
that  OR  is  equal  to  ES.  For  we  have 

OR  =  OB.cosBOR  =  OF.tan0.sin  VOS  =  VS .  tan  c/>  =  SE. 

Hence  it  follows  that  the  locus  of  R  will  be  this  smaller  semicircle, 
or  since  the  current  is  proportional  to  SE,  the  locus  of  a  vector 
drawn  from  0  to  represent  the  current  is  the  above  circle.  Again, 
since  the  angle  SVE  is  always  the  constant,  </>,  and  VSO  is  a  right 
angle,  the  external  angle  VEO  is  constant,  and  it  follows  that  E 
will  also  describe  a  semicircle  drawn  on  the  line  OD  perpendicular 
to  OQ  as  diameter,  this  circle  also  passing  through  V.  We  can 
now  see  that  the  line  QR  is  always  proportional  to  OE,  that  is  to 
the  E.M.F.  produced  by  the  rotation  of  the  armature.  For  the 
angle  EO  Y  being  the  angle  between  the  tangent  0  Y  at  0  and  the 
secant  OE  at  0  is  equal  to  the  angle  in  the  segment  EZO.  But 
EOY and  EOQ  are  supplemental,  as  are  the  angle  in  the  segment 
QXR  and  the  angle  ROQ,  hence  the  angle  in  the  segment  EZO 
and  that  in  the  segment  QXR  are  equal ;  it  follows  that  the  chords 
OE  and  QR  are  in  the  ratio  of  the  diameters  OD  and  OB,  that  is 
in  a  constant  ratio,  and  thus  the  line  QR  is  always  proportional  to 
the  E.M.F.  of  the  armature.  We  can  then  draw  the  small  semi- 
circle to  a  larger  scale,  as  in  Fig.  70,  and  use  this  to  represent  all 
the  different  quantities. 

Since  the  motor  will  always  be  working  on  fairly  low  inductions 
the  reluctance  will  be  practically  that  of  the  air  paths  only  or  will 


be  approximately  constant,  hence  the  field  will  always  be  nearly 
proportional  to  the 


current,  and  hence  the  torque  produced  will  be 


Fig.  70. 

nearly  proportional  to  the  square  of  the  current  or  to  the  square 
of  OR.  Again,  it  is  evident  that  the  speed  is  directly  as  the  E.M.F. 
of  the  armature  and  inversely  as  the  field  or  the  current,  it  is  thus 
proportional  to  the  ratio  QR/OR.  The  output  will  be  proportional 
to  the  product  of  the  armature's  E.M.F.  and  the  current  or  to 


THE   SERIES   MOTOR 


99 


QR .  OR.  If  the  line  RL  be  drawn  perpendicular  to  OB  it  will 
give  the  part  of  the  current  that  is  in  phase  with  the  pressure  and 
hence  the  input  will  be  0V .RL,  and  thus  the  efficiency  will  be 
readily  found. 

In  an  actual  motor  the  losses  are  far  from  being  proportional 
to  the  square  of  the  current  as  this  construction  implies  and  hence 
we  should  not  expect  the  semicircle  to  give  accurately  the  value 
of  the  current.  If  a  test  is  made  it  will  be  found  that  the  locus  of 
the  current  vectors  is  no  longer  a  semicircle  but  very  closely  lies 
on  an  arc  of  a  circle  which  is  somewhat  larger  or,  in  some  cases, 
somewhat  smaller  than  the  semicircle.  In  any  case  the  circle  being 
assumed  to  represent  the  facts  of  the  case  it  can  readily  be  found. 
Let  the  current,  pressure  and  power  taken  on  standstill  be 
measured,  and  deduce  the  power  factor  from  this,  which  gives  the 
angle  VOQ  in  Fig.  70.  The  corresponding  value  of  the  current 
vector  can  readily  be  found,  all  that  is  necessary  is  to  increase  the 
observed  value  of  the  current  taken  at  the  pressure  used  in  the 
ratio  that  that  pressure  bears  to  the  ordinary  working  pressure  of 


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CURVES    OF    WESTINGHOUSE    2SO   H3. 
SINGLE  PHASE  SERIES   MOTOR 


Fig.  71. 


7—2 


100 


ALTERNATING   CURRENTS 


the  machine,  and  this  will  give  the  distance  OQ.  If  the  motor  be 
loaded  to  any  desired  extent,  and  the  pressure,  which  should 
be  the  proper  working  pressure,  the  current  and  the  power  be 
measured  in  this  case,  we  evidently  have  sufficient  data  to 
determine  the  value  of  such  a  current  vector  as  OR  and  the  angle 
VOR.  Hence  the  three  points  0,  R,  and  Q  being  known,  the 
circle  can  be  at  once  drawn. 

In  Fig.  71  are  given  curves  showing  the  relation  between  the 
current  taken  by  a  series  motor  of  modern  design  and  the  brake- 
horse-power,  tractive  effort  and  speed,  efficiency  and  power  factor. 
The  tractive  effort  and  speed  refer  to  the  special  car  on  which  the 
motor  was  used,  which  was  one  for  railway  work.  The  motor  was 
designed  to  operate  at  250  volts,  and  it  will  be  seen  that  in  all 
respects  the  results  will  bear  comparison  with  a  direct  current 
motor.  In  particular,  the  mechanical  characteristics  are  as  favour- 
able for  traction  purposes  as  those  of  the  direct  current  one. 

When  traction  is  being  undertaken  by  direct  current  series 
motors  the  car  is  equipped  with  two  or  more  motors  which  can  be 
put  in  series  or  parallel.  In  such  a  case  certain  relations  exist 


/*r  POSITION      2HOPOSITIQM.          3*° POSITION.      +TH  POSITION. 


5™ POSITION 


Fig.  72. 

between  the  torque  and  speed  at  definite  pressure  applied  to  the 
system,  which  relations  depend  on  the  coupling  of  the  motors  and 
are  attained  with  no  extra  apparatus.  Such  relations  are  termed 
the  "free  running"  conditions,  and  are  such  that  no  waste  of 
energy  occurs  other  than  that  unavoidably  present  in  the  motors. 
If  it  is  desired  to  obtain  any  other  speed-torque  relations  than 
those  corresponding  to  the  free  running  conditions,  this  can  only 
be  attained  by  using  a  different  pressure  at  the  terminals  of  any 
such  combination,  and  with  direct  currents  such  diminution  of 


THE   SERIES   MOTOR  101 

pressure  can  only  be  produced  by  means  of  permitting  the  current 
on  its  way  to  the  motors  to  produce  a  drop  in  a  series  resistance. 
Hence  any  but  the  free  running  series  or  parallel  conditions 
necessitates  an  extra  waste  of  energy.  In  the  case  of  alternate 
currents  we  have  the  possibility  of  obtaining  different  terminal 
pressures  on  the  motor  by  means  of  transforming  from  the  given 
supply  pressure  to  any  other  pressure  suitable  for  the  conditions 
required  to  be  fulfilled.  For  this  purpose  it  is  usual  to  employ  an 
auto-transformer  (see  p.  80)  with  tappings  brought  out  at  such 
points  as  to  give  the  desired  pressure.  The  maximum  pressure 
used  in  motors  varies  to  some  extent,  but  in  the  motor  whose 
curves  are  given  it  was,  as  stated,  250  volts.  In  addition  to  this 
other  less  pressures  are  required  which  have  the  values  shown  in 
Fig.  72.  It  will  be  noted  that  the  tappings  are  not  brought 
direct  to  the  auto-transformer,  but  two  points  are  tapped  off  from 
it,  and  an  inductive  coil  is  bridged  across  these  points,  from  the 
centre  of  which  the  pressure  to  the  motor  is  taken  off.  This  coil 
is  called  a  "preventive"  coil,  and  is  of  use  in  damping  out  by 
inductive  action  any  short-circuit  currents  that  may  flow  in  the 
process  of  switching  from  one  tapping  to  another. 

Since  such  an  auto-transformer  gives  a  wide  and  easily  changed 
range  of  pressure,  there  is  no  necessity,  when  more  than  one  motor 
is  used  on  the  car,  to  provide  for  any  other  connection  of  the  same 
than  the  parallel  one,  it  is  however  necessary  to  provide  a  switch 
gear  to  reverse  the  connections  of  the  fields  relative  to  the 
armatures  in  order  to  provide  for  reversing  the  motion  of  the  car. 
In  general  for  heavy  traction  four  motors  of  the  type  considered 
are  used  in  parallel. 

It  should  be  noted  that  since  the  pressure  applied  is  readily 
transformed  in  any  desired  ratio,  the  actual  value  of  that  applied 
pressure  does  not  affect  the  car  equipment,  very  different  pressures 
on  the  line  can  be  utilized  by  using  auto-transformers  of  the 
proper  ratio  to  reduce  the  line  pressure  to  the  standard  pressure 
required  by  the  motors.  Further,  all  the  manipulation  is  on  the 
low  pressure  side  of  the  auto-transformer  and  the  high  pressure 
one  needs  only  a  switch  and  fuse. 


CHAPTER  IX. 


THE   E.M.F.   OF  AN  ALTERNATOR. 


E.M.F.  of  an  alternator.  The  E.M.F.  of  an  alternator  is 
always  specified  by  its  virtual  value  and  thus  depends  on  the  form 
of  the  instantaneous  curve.  For  example,  with  a  sine  wave  of 

pressure  we  know  that  £=*— ~  E.     If  the  curve  be  a  pointed  or 
triangular  one,  it  is  readily  seen  that  the  relation  is  £=—j^  E, 

\l  o 

while  with  a  rectangular  shaped  wave  we  have  £=  E,  thus  the 
same  virtual  E.M.F.  will  be  produced  with  very  different  values  of 


Flux 


I 
! 


£.  M.F. 


w 

Fig.  73. 


THE    E.M.F.    OF   AN    ALTERNATOR 


103 


the  maximum  depending  on  the  form  of  the  wave.  This  form 
varies  with  two  factors,  the  form  of  the  induction  curve,  that 
is,  the  relation  between  the  angular  position  of  the  armature 
and  the  flux  through  a  single  loop  of  the  armature,  and  the 
arrangement  of  the  different  loops  forming  a  coil  on  it,  that  is,  on 
the  nature  of  the  winding. 

Induction  curve.     Influence  of  form  of  flux.     Let  us 

first  consider  the  effect  of  the  form  of  the  induction  curve,  and 
let  the  coil  consist  of  I  loops  all  concentrated  in  one  place  rotating 


Fig.  74. 

with  uniform  velocity  in  a  uniform  field  as  in  Fig.  1.  In  this 
case,  as  we  have  seen,  the  flux  through  the  coil  will  be  a  sine 
function  of  the  time.  With  any  other  form  of  induction  curve 
the  E.M.F.  one  will  differ  from  a  sine,  and  will  be  widely  different 
from  that  of  the  induction  curve.  In  Figs.  73  to  75  are  shown 
three  assumed  induction  curves  and  the  corresponding  E.M.F.  ones 
for  a  single  loop.  In  each  case  the  ordinate  of  the  E.M.F.  curve  is 
roughly  drawn  to  be  equal  to  the  slope  at  each  point  of  the 
induction  curve.  On  referring  to  the  first  two  it  will  readily  be 
seen  that  a  flat  induction  curve  will  produce  a  pointed  E.M.F.  one 
and  vice  versa.  In  the  third  case  an  induction  curve  having  four 
different  slopes  in  the  half  period  is  taken  and  the  very  different 
E.M.F.  curve  resulting  is  seen.  The  student  should  sketch  in 
various  forms  of  induction  curve  and  deduce  the  corresponding 
E.M.F.  ones. 

In  any  one  curve  let  the  maximum  flux  through  a  single  loop 
of  the  coil  be  <f> ;  since  the  coil  consists  of  I  concentrated  loops  the 
maximum  flux  passing  through  the  armature  will  be  <!>/,  hence 


104 


ALTERNATING   CURRENTS 


when  the  coil  turns  through  a  quarter  revolution,  starting  with  its 
plane  perpendicular  to  the  flux,  the  change  of  flux  will  be  <&l. 


If  the  armature  execute  n  rotations  per  second  the  time  taken  for 
this  quarter  turn  will  be  -r-  seconds,  and  hence  the  mean  rate  of 


change  of  the  flux  will  be  ^n^l.  The  same  will  hold  for  a  half 
rotation,  and  although  any  further  rotation  will  result  in  change 
in  the  direction  of  the  flux,  we  can  say  that  in  any  given  case  the 
mean  E.M.F.  produced  will  be  given  by  the  above  expression  :  the 
curve  in  Fig.  76  shows  these  successive  additions  of  flux  in  a 
revolution.  It  is  customary  to  deal  with  the  number  of  con- 
ductors that  are  at  any  instant  in  series  instead  of  the  number  of 
loops,  and  it  is  evident  that  the  former  are  twice  the  latter,  hence 
if  T  denote  as  usual  the  conductors  that  are  in  series  we  have 
21  =  T,  the  expression  for  the  mean  E.M.F.  produced  with  the 


THE   E.M.F.   OF   AN   ALTERNATOR  105 

maximum  flux  <£,  will  be  mean  e  =  2.  <&nT.   This  must  be  true  what- 
ever the  shape  of  the  induction  curve,  and  thus  if  we  take  the  case 


Fig.  76. 

where  that  curve  is  simple  harmonic  such  as  is  produced  by  rotation 
in  a  uniform  field,  we  know  that  the  virtual  EM.F.  is    —  -     or  I'll 


times  the  mean,  hence  in  such  a  case  we  can  put  S—  2* 
The  student  will  possibly  feel  a  difficulty  at  this  point.  Let  the 
same  field  be  existent,  but  let  the  conductors  be  joined  up  so  as  to 
form  a  direct  current  armature  of  T  peripheral  wires.  We  know 
that  then  the  E.M.F.  is  given  by  E  =  <&nT,  so  that  the  alternator 
apparently  produces  2*22  times  the  E.M.F.  It  must,  however,  be 
recollected  that  in  the  latter  machine  all  the  conductors  have 
been  taken  to  be  in  series,  while  in  the  former  only  half  are  in 
series  at  any  moment,  T  being  in  both  cases  the  total  number  of 
peripheral  conductors.  In  the  direct  current  machine  the  current 
has  two  paths  inside  the  armature  and  but  a  single  one  in  the 
alternator,  hence  with  the  same  winding  the  alternator  could  only 
carry  half  the  current.  Thus  while  the  E.M.F.,  with  concentrated 
winding,  is  2'22  times  as  much  in  the  alternator,  the  current  is 
but  one  half. 

With  any  other  curve  of  magnetic  flux  than  the  sine  having 
the  same  maximum,  the  virtual  E.M.F.  produced  will  still  be 
proportional  to  4>,  n  and  T  but  the  factor  will  no  longer  be  2'22, 
we  can  however  say  that  in  any  case  S=k.  <&nT  where  the  k  has 
various  values  depending  on  the  form  of  the  induction  curve  and 
that  when  the  latter  is  a  sine  function  of  the  time,  this  factor  is  2*2. 

Influence  of  form  of  winding  :  breadth  coefficient.  The 
next  point  is  to  see  what  effect  is  produced  when  the  winding  is 


* 


field    ^  g 

Fig.  77. 


106  ALTERNATING   CURRENTS 

not  concentrated  with  all  the  loops  in  the  same  place,  but  spaced 
out  on  the  armature  core.  This  must  be  always  the  case  to  some 
extent  in  practice,  and  we  shall  see  that  for  some  reasons  such  an 
arrangement  has  special  advantages.  Consider  a  coil  of  two  loops 
rotating  in  a  uniform  field,  if  the  two  occupy  closely  the  same 
position,  and  if  e  is  the  E.M.F.  due  to  either,  the  E.M.F.  of  the  coil 
will  evidently  be  2e.  Now  let  them  be  wound  at  an  angle  6  as 
shown  in  Fig.  77 ;  in  the  figure  for  the  sake  of  clearness  the  coils 
are  shown  rotating  about  an  axis  parallel  to  their  own  axes, 
with  a  uniform  field  in  the  direction  of  the  arrow ;  this  evidently 
makes  no  difference  to  the  E.M.F.  produced,  it  will  still  be  sinu- 
soidal and  will  have  a  definite  maximum  equal  to  the  expression 
given  on  p.  2.  If  the  two  vectors  AA^  and  BBl  be  drawn 
each  of  length  equal  to  the  maximum  E.M.F.,  e,  in  either  coil 
and  making  the  angle  6  with  one  another,  the  projections  of  these 
will  give  the  corresponding  instantaneous  E.M.F.S.  If  the  two  loops 
be  now  joined  in  series  in  the  proper  direction  so  that  the  E.M.F.S 
add,  the  resultant  E.M.F.  will  have  a  maximum  given  by  the  sum 
of  the  two  vectors  as  shown  in  the  figure  at  OA  and  OB,  this 
resultant  being  00.  It  will  attain  its  maximum  at  an  angle  6/2 
before  BBl  and  the  same  angle  after  AA^  and  thus  that  maximum 
will  be  attained  when  the  constituent  loops  lie  at  that  angle  to 
the  direction  of  the  field,  or  in  other  words  when  the  resultant 
00  lies  along  that  field.  Hence  in  any  other  cases  of  this  nature 
the  position  in  which  the  compound  coil  lies  when  it  is  producing 
the  maximum  E.M.F.  must  be  such  that  the  axis  of  symmetry  of 
the  set  of  vectors  representing  the  constituent  E.M.F.S  is  along  the 
direction  of  the  uniform  field. 

In  this  case  the  resultant   of  the  two  vectors  is   evidently 
r\ 

2e .  cos  -  since  the  length  of  either  is  e.     Hence  the  spacing  of  the 

0 
two  coils  has  resulted  in  reducing  the  E.M.F.  in  the  ratio  of  2 .  cos  ^ 

r\ 
to  2.     It  follows  that  the  E.M.F.  will  be  given  by  E  =  4*4  cos    <E>. 


THE    E.M.F.    OF    AN   ALTERNATOR 


107 


Thus  if  in  any  similar  case  the  virtual  E.M.F.  in  a  constituent 
loop  is  given  by  2'2 .  4>?i  when  the  field  is  uniform,  the  E.M.F. 
due  to  the  combination  can  be  written  ^"=2*2  .6.  <&nT,  where  6 
is  a  number  less  than  unity.  This  number  is  called  the  "  breadth 
coefficient"  of  the  arrangement  and  we  will  now  calculate  it  for 
certain  arrangements  of  coils,  in  each  case  taking  the  position  of 
maximum  E.M.F.  for  the  combination,  that  is  the  symmetrical 
position  referred  to  above. 

First  take  the  case  where  the  two  loops  are  at  right  angles,  it 
will  readily  be  seen  from  Fig.  78  that  in  this  case  the  ratio  of  the 
actual  E.M.F.  to  that  which  would  have  been  produced  with 


Fig.  79. 

concentrated  windings  of  the  same  number  of  loops,  that  is  the 
breadth  coefficient,  6,  is  ~  or  0707.     The  next  figure  (79)  shows 

three  loops  at  60°  in  which  case  b  is  evidently  §  or  0'667,  while 
with  four  loops  (Fig.  80),  we  get  b  =  f  (cos  22  J°  +  cos  67£°)  or  0'653. 


Fig.  80. 


108 


ALTERNATING   CURRENTS 


The  limiting  case  will  be  when  we  have  a  uniformly  wound  coil  with 
a  distributed  winding  of  many  turns,  such  for  example  as  one  of 
the  type  of  a  Gramme  ring  (see  Fig.  81).  Consider  the  position 


Fig.  81. 

where  the  E.M.F.,  due  to  the  whole  assemblage  of  coils,  is  a 
maximum ;  the  loops  will  then  be  symmetrically  spaced  round  the 
middle  loop,  M ;  this  will  be  producing  the  maximum  E.M.F.,  e. 
Any  loop  making  the  angle  <f>  with  this  one  will  give  an  E.M.F.  , 
e  cos  <£.  Let  there  be  T  loops,  then  the  loops  in  a  belt  of 

T 

breadth  d<j>  will  be  — .  d(f>  and  the  E.M.F.  due  to  the  whole  set  of 


7T 


loops  which  occupy  half  the  circumference  will  be 

•a 

Te  p  ,  .         2Te 

-      cos  6  .  dd>  or  --  . 
7rjn  TT 

2 

If  they  were  concentrated  the  E.M.F.  would  have  been  Te.     Hence 

2 
the  value  of  b  is  —  or  0'635. 

7T 

This  last  case  would  be  realized  in  the  case  where  an  ordinary 
direct  current  armature  has  two  opposite  points  in  the  winding 
connected  to  slip  rings.  Since  only  half  the  coils  are  in  series  if 
we  still  use  T  for  the  total  surface  windings  our  expression  for  the 
E.M.F.,  assuming  a  sine  flux,  will  be  for  concentrated  windings 


—J  <&nT  as  the  coils  in  series  are  only  T/2.     Since  the 

9 

breadth  coefficient  was  found  to  be  —  ,  the  virtual  pressure  between 


THE   E.M.F.   OF  AN  ALTERNATOR  109 

the  rings   will   be    £=—r=<l>nT.     We   can   derive   this  result  in 

y  L 

another  way.  Since  the  maximum  of  the  alternating  E.M.F.  must, 
from  the  nature  of  the  case,  be  the  same  as  the  direct  current 
pressure,  it  will  be  4>nT,  hence  the  virtual  value  of  it  will  be 


This  extreme  form  of  distributed  winding  is  only  found  in  the 
case  just  considered,  where  slip  rings  are  used  in  connection  with 
a  direct  current  armature,  this  apparatus  is  called  a  Rotary 
Converter  and  will  be  dealt  with  later  on.  The  ratio  determined 
above  is  not  exactly  fulfilled  even  in  this  case  since  the  induction 
curve  is  not  a  sine  curve  in  actual  cases. 

In  some  cases  a  similar  distributed  winding  is  used  which 
embraces  less  than  180°  (Fig.  82).  It  is  easily  seen  that  if  the 


Fig.  82. 

angle  subtended  by  such  a  winding  be  denoted  by  2\jr  the  value 
of  b  is  given  by  ^-y  I       cos  <£  .  d<f>  or        ^ .     For  example  if  the 

coils  cover  a  quadrant  -^r  =  45°  and  b  becomes  0'90,  with  an  angle 
of  120°,  6  is  0'82,  while  with  one  of  60°,  b  is  0'95. 

Since  with  a  concentrated  winding  and  any  form  of  induction 
curve  we  found  that  the  E.M.F.  produced  is  &  =  k .  <&nT,  if  the 
same  winding  be  distributed  we  must  write  £=k.b.<&nT,  where 
b  is  the  breadth  coefficient.  Combining  the  two  constants  into 
one  we  get  £=K .<£>nT;  the  value  of  K  varies  from  0'6  to  2'3 
in  different  types,  being  greater  for  concentrated  than  distributed 
windings. 

Multipolar  fields.  Up  to  the  present  we  have  considered 
that  the  armature  producing  the  E.M.F.  rotates  between  a  pair  of 


110 


ALTERNATING    CURRENTS 


poles,  and  thus  one  period  is  produced  per  revolution.  In  practice 
the  periodicity  employed  varies  between  about  25  and  100  depend- 
ing on  the  circumstances  of  each  case.  Thus  to  attain  a  periodicity 
of  80  the  armature  would  have  to  revolve  at  4,800  R.P.M.  which  is 
far  greater  than  the  ordinary  speed  of  any  prime  mover  other  than 
some  forms  of  steam  turbine,  and  is  greater  than  is  desirable  for 
driving  by  belts,  etc.  It  becomes  necessary,  therefore,  to  provide 
more  than  a  single  pair  of  poles  in  the  field  magnet  in  order  that 
the  necessary  periods  may  be  produced  at  the  desired  speed  of 
the  prime  mover.  Thus  if  such  a  prime  mover  rotates  at  such 
a  speed  as  to  cause  the  dynamo  to  make  m  R.P.M.  and  if  we  require 
n  periods  per  second  it  follows  that  the  dynamo  must  possess 
a  number  of  pairs  of  poles,  p,  given  by  the  equation 


m 


Consider    such    a    crown    of    eight    poles,   or   four   pairs,   as 
in   Fig.  83,  and  let  our  coil  be  fixed  to  a  core  and  rotated  as 


Fig.  83. 

shown.  Apart  from  the  fact  that  the  curve  of  flux  into  the  coil 
is  of  necessity  no  longer  sinusoidal  with  time,  the  coil  will  experi- 
ence the  same  cycle  of  flux  changes  as  it  passes  from  one  of  the 
north  poles  in  the  crown  to  the  next  one,  as  it  would  have 
experienced  in  simply  rotating  once  between  the  two  polar  faces 
in  the  elementary  case.  Thus  if  we  call  the  distance  (whether 
angular  or  linear)  between  the  centre  lines  of  two  successive 
similar  poles,  the  pitch  of  the  poles,  we  see  that  the  coil  has  one 
period  produced  in  traversing  the  pitch,  or  that  the  pitch  corre- 


THE   E.M.F.   OF   AN   ALTERNATOR  111 

spends  to  360°  of  the  period.  The  pitch  is  sometimes  referred  to 
as  containing  360  electrical  degrees,  although  in  this  case  it 
corresponds  to  an  actual  motion  of  only  90°  in  space.  Instead  of 
having  all  the  loops  of  the  coil  wound  on  one  projection  of  the 
armature  it  would  manifestly  be  an  improvement  to  wind  the 
same  loops  in  four  symmetrically  placed  projections  as  in  Fig.  84, 
the  direction  of  winding  of  all  these  being  the  same.  In  this  case 
we  have  half  as  many  coils  as  poles  and  the  winding  is  called 
hemitropic.  If  it  is  desired  to  have  the  same  number  of  coils  as 
poles,  all  that  must  be  done  is  to  wind  eight  coils,  one  for  each 


Fig.  84.  Fig.  85. 

pole,  but  in  such  a  way  that  the  successive  intermediate  coils  are 
connected  in  the  reversed  direction  to  the  others,  since  at  the 
instant  they  are  opposite  the  south  poles  of  the  crown,  the  first  set 
is  opposite  the  north  poles,  and  in  order  that  the  E.M.F.S  produced 
in  each  coil  may  add,  the  coils  in  the  two  sets  must  cut  the  fluxes 
in  opposite  directions.  This  condition  can  be  fulfilled  in  many 
ways;  one  of  them  is  shown  in  Fig.  85.  For  full  details  of  the 
numerous  forms  of  windings  the  student  is  referred  to  any  standard 
book  on  the  subject. 

The  windings  of  the  coils  in  the  armature  is  now  usually 
carried  out  in  slots  left  in  the  laminated  armature  core.  The 
latter  is  formed  in  the  same  way  as  the  direct  current  cores  by 
assembling  thin  washers  of  soft  iron  on  a  shaft,  the  washers  having 
teeth  stamped  in  them  to  form  the  slots.  In  a  concentrated 
winding  there  would  be  one  or  two  such  slots  per  pole,  in  the 
distributed  windings  there  would  be  more  and  the  number  of  slots 
per  pole  will  be  a  measure  of  the  amount  of  distribution  adopted. 

The  crown  of  poles  is  commonly  excited  with  a  direct  current 
passing  round  suitable  bobbins  placed  on  the  poles  of  the  crown. 
In  all  but  quite  small  machines  it  is  found  that  instead  of  the 
crown  of  poles  being  fixed  and  the  armature  rotating  within  it 
the  converse  arrangement  is  advisable.  The  armature  is  a  more 
complex  affair  than  the  polar  crown  and  hence  can  with  advantage 
be  made  stationary,  furthermore  with  high  pressures  such  an 


112 


ALTERNATING  CURRENTS 


Fig.  86. 


THE   E.M.F.   OF   AN   ALTERNATOR 


113 


arrangement  is  safer  in  operation.     In  such  a  case  the  exciting 
current  is  led  into  the  field  magnets  by  a  pair  of  slip  rings. 

In  Fig.  86  is  given  a  view  of  a  complete  alternator  of  the  fixed 
armature  type,  together  with  a  portion  of  the  armature  showing 
the  arrangement  of  the  coils.  The  machine  is  actually  a  polyphase 
one,  but  the  design  of  a  monophase  alternator  would  be  much 
the  same. 

Forms  of  E.M.F.  curves  with  crown  of  poles.  We  will 
now  consider  the  form  of  E.M.F.  curve  produced  by  this  crown  of 
poles  in  a  few  special  cases.  Take  first  that  in  which  the  space 

K Pitch  -  360' *| 


N 

S 

„ 

I        [ffllllllllHIIIIIIIIIIIIIDJ 

4-  I 


loiiiiiiiiiiiiiniiinpj 


Angular  position   of   Coil. 


Fig.  87. 

occupied  by  a  pole  and  the  space  between  them  are  the  same 
in  amount,  and  the  flux  comes  out  uniformly  all  along  the  pole 
without  any  fringing,  such  a  case  as  is  shown  in  Fig.  87.  For 

L.  8 


114 


ALTERNATING   CURRENTS 


the  sake  of  simplicity  the  poles  are  drawn  on  a  straight  line  base, 
and  the  pitch  marked  corresponds  to  one  complete  period.  Let 
the  armature  loops  be  concentrated  and  have  the  same  pitch 
as  the  poles,  so  that  each  loop  just  catches  the  full  flux  from  a  pole 
when  it  is  opposite  to  it.  The  relation  between  the  flux  in  the 
coils  and  their  position,  that  is,  the  induction  curve,  will  be  as  in 
the  top  curve.  It  follows  that  at  constant  angular  velocity  the 
relation  between  that  position  and  the  E.M.F.  in  any  loop  will 
be  as  shown  in  the  lower  curve  since  the  E.M.F.  is  the  change 
rate  of  the  flux.  Let  the  coil  have  three  concentrated  loops, 
then  the  E.M.F.  will  be  the  same  shape  as  the  above  curve  of 
E.M.F.  but  the  ordinates  will  be  three  times  as  big.  Now  let 
the  same  loops  be  placed  at  positions  distributed  along  the 
armature ;  each  loop  will  give  its  appropriate  E.M.F.  of  the  same 
shape  as  the  original  curve  of  E.M.F.  and  the  same  height,  but  they 
will  be  displaced  laterally,  this  is  shown  in  Fig.  88,  where  the 


1     3 

rn*K 

2 

Y| 

r  1  i  Z  I  3 

2\1\ 

J*l'i 

2.  1  / 

1 

J2  {  3 

2}  1  \ 

1  \2\3 

V.         L        J 

Three  Loops 
Distributed 


Fig.  88. 

three  curves  are  shown  in  full  and  dotted  lines.  The  numbers  on 
the  diagram  show  how  many  of  these  constituent  curves  are  lying 
over  one  another.  If  the  three  curves  be  added,  noting  carefully 
that  the  parts  above  the  axis  are  positive  and  those  below 
negative,  we  get  the  resultant  E.M.F.  curve  for  the  distributed 
winding  as  shown  in  the  figure.  It  is  evident  that  while  the 
maximum  in  this  case  is  the  same  as  for  the  concentrated  winding 
the  virtual  value  of  the  E.M.F.  is  much  less.  The  effect  of  any 
fringing  at  the  sides  of  the  poles  is  to  smooth  out  the  angles 
of  the  above  curves  and  thus  give  more  regular  outlines.  It 
should  be  noted  that  the  distribution  of  the  loops  results  in  more 
gradual  variation  in  the  E.M.F.,  in  fact  that  the  resultant  curve  more 
nearly  approaches  a  sine  one  than  any  of  the  components.  This 
will  be  again  referred  to  later  on. 

Harmonics.     The  simplest  possible  form  of  alternating  quan- 
tity is,  as  we  have  said,  one  whose  trace  is  the  curve  of  sines,  or 


THE   E.M.F.    OF   AN    ALTERNATOR 


115 


the  simple  harmonic  curve.  It  can  be  shown  that  any  other 
alternating  quantity  whatever  can  be  considered  to  be  made  up  of 
a  set  of  such  quantities  of  different  amplitudes  and  having 
frequencies  which  are  simple  multiples  of  that  of  the  alternating 
quantity.  Each  member  of  the  whole  series  of  such  sinusoidal 
quantities  is  known  as  an  harmonic  of  the  alternating  quantity, 
the  one  having  the  same  period  as  the  quantity  being  called  the 
fundamental  and  the  others  being  called  the  second,  third,  fourth 
etc.  harmonics.  It  will  be  seen  that  they  fall  into  two  sets,  those 
having  an  odd  number  of  times  the  period  of  the  fundamental,  and 
those  having  an  even  number  of  times  that  period.  The  former 


Fig.  89. 

(including  the  fundamental)  are  called  the  odd  harmonics,  the 
latter  the  even  ones. 

Consider  the  case  shown  in  Fig.  89  where  a  fundamental  has 


Fig.  90. 


8—2 


116 


ALTERNATING   CURRENTS 


been  combined  with  its  second  harmonic,  the  two  starting  at  zero 
together.  At  the  first  zero  point  of  the  fundamental  the  two 
harmonics  increase  together,  while  at  the  second  zero  point  they 
increase  in  opposite  directions;  it  follows  that  the  form  of  the 
resultant  curve  as  it  rises  from  one  zero  value  is  different  from  its; 
form  when  rising  negatively  from  the  other,  and  further  it  is 
evident  that  this  result  is  true  for  all  the  even  harmonics.  Now 
in  any  alternating  pressure  or  current  it  must  follow,  from  con- 


Fig.  91. 

ditions  of  symmetry,  that  the  form  of  the  curve  as  it  is  increasing 
in  value  from  zero  cannot  depend  on  the  direction  in  which  it  is 
changing,  and  hence  no  even  harmonics  can  occur  in  the  ordinary 


Fig.  92. 


THE   E.M.F.   OF   AN   ALTERNATOR  117 

cases  of  alternating  pressures  or  currents.  Now  take  the  cases 
shown  in  Figs.  90  to  92,  where  the  fundamental  is  taken  with  its 
third  harmonic.  The  three  cases  show  the  third  harmonic  having 
three  different  phase  relations  at  starting;  in  Fig.  90  the 
fundamental  and  the  harmonic  rise  together  from  zero,  in 
Fig.  91  the  harmonic  rises  oppositely  to  the  fundamental  or  is 
antiphased  at  the  start,  and  in  Fig.  92  the  harmonic  does  not 
attain  its  zero  till  the  fundamental  has  gone  through  30°  of  the 
full  period.  It  will  be  seen  that  very  different  forms  of  curve 
result  with  the  same  harmonics  depending  on  the  relative  phase  of 
the  two.  It  follows  that  the  expression  for  the  compound  curve, 
in  addition  to  the  amplitudes  of  the  harmonics,  must  contain  the 
relative  phases.  The  mathematical  expression  for  the  above 
cases  will  be  respectively 

y  =  A!  sin  x  +  As  sin  3#,     y  =  Al  sin  x  —  As  sin  3x, 
and  = 


where  <f>3  is  the  angle  shown,  x  is  used  for  the  independent  variable 
instead  of  time  to  save  writing.  The  whole  abscissa  of  the 
fundamental  will  be  360  in  degrees  or  the  periodic  time  T  in 
seconds. 

If  the  curve  be  more  complicated  in  form  it  is  necessary  to 
include  more  harmonics  than  the  third,  each  with  its  appropriate 
angle  of  lag,  and  the  complete  expression  for  any  curve  whatever 
will  be 

y  =  A-L  sin  x  +  A3  sin  3  (x  —  <f>3)  +  A&  sin  5  (x  —  </>5)  +  etc. 

By  assigning  the  proper  values  to  the  amplitudes  and  phase 
angles  this  expression  can  be  made  to  represent  any  assigned 
alternating  quantity.  In  most  cases  of  E.M.F.S  and  currents  we 
rarely  want  higher  harmonics  than  the  seventh,  though  in  certain 
circuits  the  effect  of  still  higher  ones  has  to  be  taken  into  account. 

Effect   of  harmonics   on   current   curve.      Let   us   now 

consider  the  effect  of  the  higher  harmonics  on  circuits  of  the 
ordinary  type.  For  example  let  the  E.M.F.  be  given  by 

e  =  E!  sinpt  +  E3  sin  3pt  +  Eg  sin  5pt, 

and  let  it  send  a  current  through  a  circuit  of  resistance  R  and 
self-induction  L.  Each  harmonic  in  the  E.M.F.  will  produce  its 
appropriate  current.  Thus  the  fundamental  will  give  the  current 

Cl  =  (7 

where 


118 


ALTERNATING   CURRENTS 


The  two  higher  harmonics  will  give  the  currents 


and 


where 


tan  X,  = 


__3pL 


R 


and    tan  X5 


5pL 
R 


The  total  current  will  be  the  sum  of  these  three  currents.  It 
follows  that  in  this  case  the  harmonics  are  much  less  evident  in 
the  current  curve  than  in  the  E.M.F.  one,  and  that  the  harmonics 
are  altered  in  phase,  hence  the  current  curve  is  very  different  in 
shape  from  the  E.M.F.  one.  In  the  case  of  a  condenser  being 
supplied  by  the  same  E.M.F.  the  three  currents  will  be 

Cj  =  El.pF  .cospt, 
cs  =  3.E3.pF.  cos  3pt  and  c5  =  5  .  E5.pF.  cos  5pt. 

Thus  in  this  case  the  amplitude  of  the  harmonics  will  be  increased 
in  the  current  curve.  It  will  be  recollected  that  when  both 


Fig.  93. 


THE    E.M.F.    OF   AN    ALTERNATOR  119 

capacity  and  self-induction  are  present  the  phenomenon  of  re- 
sonance occurs  for  certain  relations  between  the  values  of  these 
quantities  and  the  periodicity,  it  follows  that  with  a  compound 
curve  of  pressure,  resonance  may  occur  for  any  of  the  periods 
corresponding  to  any  of  the  harmonics.  Thus  in  the  present  case 
resonance  may  occur  for  the  three  relations 

LFp*  =  1,     9  .  LFp*  =1,     25  .  LFp*  =  1  ; 

where  p  denotes  27m  for  the  fundamental,  and  in  more  compli- 
cated cases  a  further  number  of  similar  expressions  will  hold. 
Thus  when  waves  of  non-sinusoidal  form  are  used  the  chances  of 
resonance  are  increased.  In  modern  systems  the  constants  of  the 
supply  circuit,  including  mains  etc.,  is  such  that  resonance  for 
harmonics  of  other  than  fairly  high  period  is  not  likely  to  occur, 
and  even  then  the  current  due  to  the  harmonic  is  kept  within 
reasonable  limits  by  the  resistance  of  the  circuit. 

Fig.  93  shows  the  current  curve  in  the  case  where  the  con- 
ditions of  the  circuit  were  such  as  to  make  even  the  13th  harmonic 
of  some  importance,  resonance  having  occurred  for  that  harmonic 
under  certain  circumstances. 

Virtual  value  of  a  complex  quantity.  It  is  often  impor- 
tant to  find  the  virtual  value  of  such  a  quantity  as  we  have  been 
considering.  Let  y  denote  the  instantaneous  value  of  the  same 
and  ylt  y3,  and  y5  etc.  the  corresponding  instantaneous  values  of 
its  harmonics  :  then  we  have,  summing  over  a  period, 


2  fT 
-     (Mfc 

TJo 


But  the  latter  integral  is  zero  since  the  integral  of  the  product  of 
two  sine  quantities  of  different  commensurable  periodicities  is  zero 
over  a  period.  Thus  if  &  is  the  virtual  value  of  any  E.M.F.  and 
^i,  <^3»  &s  etc.  the  virtual  values  of  its  harmonics,  by  taking  means 
on  both  sides  of  the  above  we  evidently  have 


It  will  be  seen  that  this  virtual  value  is  independent  of  the 
phase  angles  of  the  harmonics. 

This  fact  is  also  evident  from  the  consideration  that  energy  is 
not  a  vector  quantity,  and  hence  the  heat  produced  by  any  complex 
current  in  a  definite  resistance  is  merely  the  arithmetic  sum  of  the 
heat  produced  by  each  separate  component.  From  this  it  follows 
immediately  that  the  square  of  the  virtual  value  of  the  -complex 


120  ALTERNATING   CURRENTS 

current  is  the  sum  of  the  squares  of  the  virtual  values  of  each  of 
its  components. 

Power  due  to  the  harmonics.  The  fact  that  power  is 
a  scalar  quantity  further  leads  to  the  following  result.  Let  the 
virtual  values  of  the  different  harmonics  in  the  pressure  curve  be 
£\>  &s>  £5  etc.  and  those  in  the  current  curve  be  ^i,  ^3,  ^5  etc., 
further,  let  the  phase  angles  between  these  several  harmonics  be 
\,  ^s,  XB  etc.  It  would  evidently  be  exactly  the  same  if  the  power 
were  being  supplied  by  a  set  of  machines  all  rigidly  geared  and 
compelled  to  move  at  the  proper  relative  speed  and  phase,  and  if 
each  produced  the  appropriate  current  and  pressure,  hence  the 
total  power  will  be  given  by 

W  =  £$i  cos  \!  +  <^8  cos  X3  +  £8<&6  cos  X5  +  etc. 
Hence  each  harmonic  is  only  productive  of  power  with  its  corre- 
sponding harmonic  in  the  current  curve.  Let  the  total  power  be 
the  same  in  amount,  but  due  to  the  passage  of  the  equivalent 
virtual  current  <$  passing  under  -the  equivalent  virtual  pressure  £t 
these  being  determined  as  just  described.  Then  this  power  will  be 
given  by  £<$  cos  X  where  X  is  the  angle  of  phase  difference  for 
these  equivalents  and  cos  X  is  the  true  power  factor.  It  evidently 
follows  that  this  power  factor  can  be  written 

cos  X  =  2  .  £n<@n  cos 


It  is  hence  evident  that  the  power  factor  for  a  non-sinusoidal  load 
must  vary  with  the  values  and  relative  phases  of  the  constituent 
harmonics  and  hence  can  only  be  considered  as  the  cosine  of  a 
definite  phase  angle  when  the  equivalent  virtual  currents  and 
pressure  are  taken  in  the  expression. 

Effect  of  harmonics  on  a  transformer.     Form  Factor. 

The  core  loss  in  a  transformer  has  been  shown  to  be  given  by 

a  .  n  .  B1*  +  b.ri>.B\ 

where  a  and  b  are  constants  depending  on  the  quality  of  the  iron 
and  the  form  and  nature  of  the  iron  core,  while  B  is  the  maximum 
of  the  induction  in  a  cycle  and  n  is  the  number  of  periods  per 
second.  Hence  at  constant  periods  the  loss  will  be  dependent  on 
the  maximum  induction  attained  by  the  iron. 

If  £  denote  the  virtual  value  of  the  applied  pressure  and 
Em  denote  its  mean  value,  we  evidently  have 


/> 


where  T  is  the  periodic  time ;  hence  /  is  a  constant  for  the  given 
form  of  E.M.F.  curve  and  is  called  its  Form  Factor.     Again  if  8  is 


-       edt 


THE    E.M.F.    OF   AN   ALTERNATOR  121 

the  section  of  the  iron  and  T  the  turns  on  the  primary  coil  the 
maximum  flux  passing  through  the  core  will  be  <£  =  BST. 

Further  if  the  flux  is  changing  the  change  of  flux  in  half 
a  period  will  be  2<I>.     But  if  n  is  the  number  of  periods  and  time 

taken  for  this  change  is  ^-  seconds,  the  mean  rate  of  change  will 

2ifi 

be  4  .  4>  .  n,  further  when  the  ohmic  drops  are  small  this  must  be 
nearly  equal  to  the  mean  value  of  the  applied  pressure  Em.  It 
follows  that  if  we  write  as  usual  p  —  %7rn  the  relation  between  the 
virtual  pressure  on  the  primary  and  the  maximum  induction  in 
the  core  will  be  given  by 

or  B  =  E_*_ 


Thus  the  maximum  induction  will  vary  inversely  as  the  value  of 
f  when  the  applied  pressure  has  the  same  virtual  value.  For  a  sine 
curve  we  have  seen  that  the  value  of  /  is  I'll  ;  for  a  completely 
flat  curve  its  value  is  evidently  unity,  and  for  a  pointed  curve 
its  value  will  be  greater  than  for  a  sine  curve.  It  follows  that 
such  pointed  curves  will  produce  less  hysteretic  loss  than  the 
ordinary  sine  curve.  Thus  for  a  curve  for  which  /  is  1*4  it  will 
readily  be  seen  that  the  induction  is  about  0'8  of  the  value  for  the 
sine  curve  and  hence  the  loss  will  be  less  than  80  °/o  of  the  loss 
with  such  a  curve. 

It  must  not,  however,  be  assumed  that  such  forms  of  curve  are 
necessarily  the  best  when  all  the  circumstances  are  taken  into 
consideration,  as  will  be  evident  from  p.  119.  Further  the  above 
result  is  only  true  on  the  assumption  that  the  hysteretic  loss,  h, 
can  be  represented  by  an  expression  of  the  form  h  =  r)(3e,  where  77 
and  e  are  constants  for  the  iron.  If  this  is  not  the  case,  the 
actual  effect  of  the  shape  of  the  E.M.F.  curve  might  be  different 
from  the  above  result.  It  is  known  that  e  is  not  a  constant 
over  the  whole  range  of  induction  commonly  used  in  alternate 
current  work,  and  hence  the  actual  effect  of  the  form  of  the 
pressure  curve  must  in  fact  be  determined  by  experiment. 

Effect  of  distributed  windings.  As  an  example  in  the  use 
of  harmonics  we  will  investigate  the  case  of  the  E.M.F.  of  an 
alternator  with  flat  topped  E.M.F.  curve  for  each  loop  which  was 
considered  on  p.  114.  It  can  be  shown  that  for  this  curve  the 
analytical  representation  of  the  ordinate  is 

4 
y  =  —  E  (sin  x  -f  J  sin  3x  +  1  sin  5#  +  etc.), 

7T 

where  E  is  the  constant  value.  In  Fig.  94  are  given  three  curves 
showing  how  the  addition  of  the  first  three  terms  approximates 
more  and  more  closely  to  the  flat  topped  wave.  For  the  sake  of 


122 


ALTERNATING    CURRENTS 


simplicity  let  us  take  the  E.M.F.  produced  in  one  of  the  loops  as 
being  given  by 

e  —  sin  x  -\-  -J-  sin  3#  -f  £  sin  5%, 

x  being  in  degrees  of  the  period.  For  five  such  loops  in  series 
placed  in  a  concentrated  form  the  E.M.F.  will  be  just  five  times  as, 
great,  and  thus  the  harmonics  will  be  present  in  the  ratio 

1  :  0-33  :  0'2. 

Let  the  five  loops  be  still  in  series  but  distant  successively  by 
15  electrical  degrees,  thus  forming  a  distributed  winding  of  five 


Fig.  94. 

loops.     Let  the  centre  one  be  taken  as  the  coil  of  reference,  then 
its  E.M.F.  will  be  given  by 


sn 


sn 


e3  =  sn  x  -4- 

The  E.M.F.S  in  the  two  loops  to  the  right  and  left  respectively 
will  be 

e2  =  sin  (x  +  15°)  +  1  sin  3  (as  +  15°)  +  1  sin  5  (x  +  15°), 
e,  =  sin  (x  -  15°)  +  £  sin  3  (x  -  15°)  +  £  sin  5  (x  -  15°), 
while  those  in  the  two  outer  loops  will  be 

e1  =  sin  (x  +  30°)  +  £  sin  3  (x  +  30°)  +  £  sin  5  (x  +  30°), 
and  e5  =  sin  (x  -  30°)  +  £  sin  3  (x  -  30°)  +  1  sin  5  (a?  -  30°). 
Considering  #2  and  e4  together  we  have 

^2  4-  04  =  2  sin  #  cos  15°  +  f  sin  3#  cos  45°  +  f  sin  5#  cos  75°, 
similarly  the  outer  two  coils,  el  and  e5,  give 

<?!  +  e5  =  2  sin  x  cos  30°  +  f  sin  3x  cos  90°  +  f  sin  5^  cos  150°. 

The  complete  E.M.F.  for  the  whole  five  in  series  will  be  the  sum 
of  the  above,  and  is 

e  =  (1  +  2  cos  15°  +  2  cos  30°)  sin  x 

+  1  (1  +  2  cos  45°  +  2  cos  90°)  sin  Sac 

+  i  (1  +  2  cos  75°  +  2  cos  150°)  sin  50?. 


THE   E.M.F.   OF  AN   ALTERNATOR  123 

This  leads  to 

e  =  4'66  sin  x  -j-  0*803  sin  3x  —  0'214  sin  5#. 
Hence  in  the  compound  curve  the  harmonics  are  in  the  ratio 

1  :  017  :  0-046, 

and  thus  we  see  that  the  distribution  of  the  winding  results  in 
a  nearer  approximation  to  the  sine  curve.  Of  course  the  virtual 
value  of  the  E.M.F.  is  diminished.  In  the  concentrated  case  it  is 

1      

-^  V  52  +  ( j)a  +  (f)2   or   3'8  volts, 

while  in  the  distributed  one  it  is 

4,  V4-662  +  0-8032  +  0-2142  or   3'35  volts. 
Y 2 

So  that  the  greater  approximation  to  the  sine  form  is  only 
obtained  at  the  expense  of  a  loss  of  virtual  pressure. 

It  follows,  then,  that  with  an  assigned  form  of  induction  curve  the 
relative  importance  of  the  harmonics  can  be  altered  by  adjustment 
of  the  winding  and  thus  by  proper  precautions  the  form  of  the 
curve  of  E.M.F.  can  be  made  very  closely  approximating  to  a  sine 
curve.  Further  change  can  be  effected,  if  necessary,  by  altering 
the  form  of  the  induction  curve  itself,  for  example  by  varying  the 
amount  of  the  air  gap  along  the  polar  face.  It  is  possible  in  this 
way  to  largely  diminish  the  amplitude  of  any  harmonic  which  it 
may  be  desired  to  eliminate  from  the  E.M.F  curve. 

It  will  be  seen  when  the  effects  of  the  armature  current  are 
under  consideration  that  these  tend  to  alter  the  shape  of  the  flux 
curve,  and  hence  to  alter  the  form  of  the  instantaneous  E.M.F. 
Such  alteration  will  in  general  tend  to  cause  deviation  from  the 
simple  harmonic  form. 

INSTANTANEOUS   CURVES. 

Point  to  Point  Method.  The  problem  of  observing  the 
instantaneous  curves  of  currents  or  pressures  has  received  much 
attention,  but  we  will  only  describe  two  out  of  the  many  methods 
that  have  been  proposed,  the  first  being  the  original  one  due  to 
M.  Joubert.  On  the  shaft  of  the  dynamo  providing  the  pressure 
are  keyed  two  carefully  turned  discs,  one  is  made  of  brass  and  the 
other  is  made  of  ebonite,  the  two  are  rigidly  fixed  together,  and 
at  one  point  a  thin  slip  of  brass  projects  from  the  brass  disc  into 
the  other  as  shown  in  Fig.  95.  On  these  two  discs  press  two  brushes 
which  are  carried  by  an  arm  capable  of  being  placed  at  any  required 
position  relative  to  the  magnets  of  the  dynamo  and  fixed  there, 
its  relative  position  being  shown  by  a  pointer  moving  over  a  scale 


124 


ALTERNATING   CURRENTS 


as  shown.  Any  apparatus  connected  to  these  brushes  will  have 
its  circuit  made  once  per  revolution.  For  example  let  this  circuit 
consist  of  the  dynamo  terminals  and  an  electrostatic  voltmeter, 


Direction  of  Rotation. 

Fig.  95. 

then  the  latter  will  indicate  the  pressure  that  the  dynamo  is 
producing  at  the  instant  the  brushes  are  joined  by  the  slip  of  brass. 
By  moving  the  arm  carrying  the  brushes  this  can  be  made  any 
point  we  please,  and  thus  from  the  scale  provided  for  the  pointer 
on  the  brush  arm  the  relation  between  the  angle  and  the  pressure 
can  be  found ;  this  will  be,  at  constant  speed,  the  same  as  the  curve 
connecting  time  with  E.M.F.,  or  in  other  words  the  instantaneous 
curve  of  E.M.F.  of  the  dynamo. 

In  many  cases  the  best  voltmeter  to  employ  is  a  suitably 
arranged  quadrant  electrometer.  One  point  should  be  noted :  it 
is  only  for  a  small  fraction  of  a  second  that  the  pressure  is  applied 
to  the  electrometer  and  for  the  rest  of  the  rotation  of  the  disc  the 
two  brushes  are  resting  the  one  on  the  brass  disc  the  other  on  the 
ebonite  one.  During  all  this  time  the  charge  of  the  electrometer 
can  leak  away  across  the  surfaces  of  the  ebonite  and  thus  the 
reading  will  be  too  small :  the  effect  of  this  can  be  made  negli- 
gible if  a  condenser  be  put  in  parallel  with  the  electrometer  so  as 
to  increase  the  charge  that  is  stored  in  the  circuit. 

The  measurement  of  the  instantaneous  values  of  a  current  can 
evidently  be  made  if  the  current  be  passed  through  a  known  non- 
inductive  resistance  and  the  curve  of  terminal  pressure  on  that 
resistance  determined  in  the  manner  just  described. 

If  a  suitable  electrostatic  instrument  is  not  available  the 
following  modified  method  may  be  employed.  Let  the  single 
brush  that  presses  on  the  ebonite  ring  be  replaced  by  two  brushes 


THE   E.M.F.    OF   AN    ALTERNATOR 


125 


fixed  to  a  bar  of  ebonite  (Fig.  96),  let  one  of  these  be  connected  to 
the  source  of  pressure  that  is  to  be  investigated,  whether  the  E.M.F. 
of  the  machine  or  some  part  thereof,  or  the  pressure  on  the  ends  of 
a  resistance,  and  let  the  other  be  connected  to  a  sensitive  deadbeat 


Fig.  96. 

galvanometer  G.  The  other  ends  of  the  two  are  connected  to  a 
condenser  F,  the  free  terminal  of  which  goes  to  the  brush  pressing 
on  the  brass  ring.  It  will  be  seen  that  when  the  strip  that 
projects  through  the  ebonite  ring  touches  one  brush  the  condenser 
is  charged  and  is  immediately  discharged  when  the  strip  touches 
the  next  brush  through  the  galvanometer.  The  successive  im- 
pulses thus  given  to  the  coil  of  the  latter  result  in  a  steady 
deflection,  since  they  occur  with  a  period  far  quicker  than  its 
natural  one.  The  calibration  can  be  effected  by  placing  a  known 
steady  E.M.F.  across  V,  the  machine  being  kept  running  at 
the  usual  speed.  For  accurate  work  care  should  be  taken  that 
the  condenser  is  fully  charged  and  discharged  each  contact,  and 
this  can  be  secured  by  making  the  strip  somewhat  broader  than 
usual,  the  pressure  measured  will  be  that  existing  at  the  instant 
the  brush  leaves  the  strip.  Since  the  action  of  the  galvanometer 
is  not  ballistic,  the  ordinary  method  of  shunting  to  secure  different 
sensibilities  can  be  used,  as  no  question  of  variation  of  damping 
can  arise  in  this  case,  the  deflection  being  a  steady  one. 

In  the  direct  methods  difficulty  is  often  met  with  from  the 
necessarily  restricted  range  of  the  electrostatic  voltmeter  or 
galvanometer  available.  The  following  null  method  avoids  this. 
In  Fig.  97  let  the  load,  L,  be  placed  in  series  with  a  known 
non-inductive  resistance,  J^,  so  that  the  current  passes  through 
both.  A  battery  whose  E.M.F.  is  somewhat  greater  than  the 
maximum  of  the  alternate  current  E.M.F.  that  has  to  be  measured 
is  connected  to  a  series  resistance,  r,  and  to  a  potential  slide, 
MN.  If  this  cannot  be  obtained,  the  alternate  pressure  must 
be  reduced  to  the  necessary  amount  by  means  of  a  suitable 
potential  divider.  By  means  of  adjusting  the  resistance  r  and 
the  position  of  the  sliders,  M  and  N,  any  desired  pressure  can 


126 


ALTERNATING   CURRENTS 


be  obtained  between  M  and  N  which  can  be  measured  on  the 
voltmeter  V.  The  points  M  and  N  are  taken  to  a  reversing  key 
Kl)  the  other  terminals  of  which  are  connected  in  series  with 


Fig.  97. 

the  alternate  pressure  to  be  measured,  the  contact  maker  C, 
two  terminals  of  a  three-way  plug  key  v,  p,  and  a  sensitive 
galvanometer  G.  It  will  readily  be  seen  that  by  suitably  adjusting 
the  sliders,  etc.,  the  pressure  applied  by  the  battery  can  be  made 
equal  to  the  instantaneous  pressure  due  to  the  alternate  current, 
and  that  condition  will  be  shown  by  the  galvanometer  G  showing 
no  deflection ;  the  value  of  the  pressure  is  read  directly  on  V.  To 
measure  the  current  it  is  passed  through  the  resistance  R1}  and  a 
second  resistance  R2  is  placed  in  series  with  R^  as  shown,  this 
may  have  the  same  value  as  Rlt  but  in  any  case  the  ratio  of  the 
two  must  be  found  accurately,  and  each  must  be  capable  of 
carrying  its  proper  current.  The  second  strip,  R2,  carries  a  direct 
.current,  supplied  by  a  distinct  and  insulated  battery,  which 
can  be  adjusted  by  the  resistance  P,  and  measured  by  the 
ammeter  A.  The  direction  of  this  current  can  be  reversed 
by  the  key  Kz.  It  will  be  seen  that  when  the  terminals  p  and  a 
are  joined,  the  differences  of  pressure  existing  between  the  two 
strips  Rl  and  R2  are  opposed  on  the  circuit  formed  by  the  contact 
maker  and  the  galvanometer.  Hence  by  adjustment  of  the  current 
shown  by  A  the  galvanometer's  deflection  can  be  made  zero,  in 
which  case  the  value  of  the  alternating  current  at  the  instant 
given  by  the  position  of  the  contact  maker  can  be  directly  read 
on  A.  Instead  of  a  single  contact  the  double  contact  method 


THE   E.M.F.    OF   AN    ALTERNATOR 


127 


of  Fig.  96  can  be  employed.  This  has  the  advantage  of  greatly 
diminishing  difficulties  due  to  leakage  currents  from  the  balancing 
batteries  if  their  insulation  is  not  very  high. 

The  Oscillograph.  The  method  just  described  is  called  the 
point  to  point  method,  and  has  the  advantage  of  giving  large 
readings,  it  is,  however,  somewhat  tedious,  and  since  each  curve 
takes  some  few  minutes  to  find  it  is  manifestly  unsuitable  for 
investigating  cases  in  which  the  phenomenon  only  occupies  a  few 
alternations.  For  such  purpose  an  instrument  known  as  the 
oscillograph  is  used.  We  will  describe  the  form  due  to  Mr 
Duddell. 

This  instrument  (Fig.  98)  essentially  consists  of  a  D'Arsonval 
galvanometer  with  a  very  light  coil  and  a  very  strong  field 
magnet.  In  order  that  the  deflections  in  such  an  instrument  may 


Fig.  98. 

accurately  follow  the  current  flowing  at  each  instant,  the  moving 
coil  must  have  a  natural  periodic  time  which  is  many  times  smaller 
than  the  impressed  period  of  any  harmonic  in  the  current  wave  that 
is  flowing  through  it,  that  is,  it  must  be  possessed  of  very  small 
inertia  and  have  a  very  great  controlling  force.  The  former 
condition  is  secured  by  making  the  coil  consist  of  two  very  thin 
and  light  strips  of  phosphor-bronze  s,  s  as  shown  in  Fig.  99. 


128 


ALTERNATING   CURRENTS 


These  strips  are  placed  close  together  in  the  air  gap  of  the  magnet 
and  a  small  piece  of  light  mirror  glass  M  is  stuck  to  them  at  the 
centre.  When  a  current  flows  in  the  strips  one  is  sucked  inwards 
and  the  other  forced  outwards  so  that  the  little  mirror  is  tilted. 
The  angle  of  tilt  is  small  and  is  nearly  proportional  to  the  current 
flowing.  The  controlling  force  is  provided  by  the  resolved  compo- 
nent of  tension  in  the  strips,  which  tension  is  given  through  a  pulley 
P  held  up  by  a  screw  carrying  a  spring  balance,  the  latter  indi- 
cating the  tension ;  this  is  made  as  high  as  is  consistent  with  safety. 
By  this  means  the  natural  period  of  the  system  is  reduced  to  less 
than  1/10,000  of  a  second.  The  space  in  which  the  strips  lie  is  in 


Fig.  99. 

addition  filled  with  oil  of  such  a  viscosity  as  to  cause  the  motion 
to  be  just  deadbeat.  Such  a  coil,  as  it  consists  of  but  a  single 
turn,  will  manifestly  need  a  very  intense  field  in  the  instrument  in 
order  that  fair  sensitiveness  may  be  attained.  This  is  secured  by 
providing  the  field  by  means  of  an  electromagnet,  which  is 
generally  designed  to  be  excited  with  some  convenient  pressure 
such  as  100  volts.  In  many  cases  it  is  desirable  to  measure 


THE   E.M.F.    OF   AN   ALTERNATOR 


129 


simultaneously  two  related  quantities  such  as  the  pressure  and 
the  current.  In  such  a  case  two  similar  coils  are  placed  in  the  air 
gap  as  shown  in  Fig.  98. 

We  must  now  consider  the  optical  arrangements.  The  source 
of  light  has  to  be  very  intense  and  is  provided  by  an  arc  lamp :  the 
beam  is  parallelized  and  passes  through  a  slit  and  a  cylindrical 
lens ;  it  then  passes  on  to  the  galvanometer  mirror  and  on  reflection 
is  met  by  a  plane  mirror,  which  in  turn  reflects  it  vertically 
on  to  a  screen  where  it  can  be  observed.  If  this  mirror  were  at 
rest  the  passage  of  an  alternating  current  in  the  coil  would  merely 
spread  out  the  spot  of  light  into  a  line,  but  if  the  mirror  is  given 
an  oscillatory  motion  the  line  will  be  turned  into  a  curve  :  in  order 
that  the  curve  may  be  that  of  the  current  in  the  strip  considered 
as  a  function  of  the  time,  we  must  arrange  matters  so  that  (1)  only 
the  forward  movement  of  the  mirror  is  used  to  reflect  the  light  in 


Fig.  100. 

order  to  avoid  confusion,  the  return  motion  taking  place  when  the 
beam  is  cut  off,  (2)  that  the  mirror  is  rocked  at  a  speed  equal  to 
half  the  number  of  periods  in  order  that  only  one  period  may  be 
visible  at  a  time,  (3)  that  the  distance  the  spot  moves  through  on 
the  screen  is  proportional  to  the  time.  These  conditions  are 


130 


ALTERNATING   CURRENTS 


provided  for  as  follows :  the  source  of  motion  of  the  mirror  is  a 
small  motor  which  is  of  the  type  known  as  "  synchronous,"  and  is 
driven  by  the  same  source  of  current  that  is  being  used  for  the 
test ;  it  is  so  designed  as  to  rotate  at  a  number  of  revolutions  per 
second  equal  to  half  the  number  of  periods.  A  screen  is  driven  by 
this  motor  (see  Fig.  100)  which  in  rotating  cuts  off  the  beam 
during  half  the  time  of  a  rotation ;  the  mirror  is  driven  by  a  cam 
which  is  so  shaped  that  the  third  condition  is  fulfilled.  Thus  on 
the  screen  will  be  seen  a  spot  of  light  which  moves  so  that  the 
abscissae  are  proportional  to  the  time  while  the  ordinates  are 
proportional  to  the  current  in  the  strip.  The  general  arrangement 
is  shown  in  Fig.  101. 


Fig.  101. 

In  the  case  where  it  is  desired  to  simultaneously  measure  the 
current  and  the  terminal  pressure  on  any  apparatus  the  oscillo- 
graph should  be  connected  up  as  shown  in  Fig.  102.  The  current 
on  its  way  to  the  load,  L,  passes  through  a  non-inductive  resistance 
jRa  on  which  one  of  the  oscillograph  strips  C  is  placed  in  parallel, 


Fig.  102. 

being  itself  in  series  with  a  second  resistance  R2.  The  two 
resistances  are  adjusted  so  that  by  tests  made  with  steady  currents, 
the  reading  of  the  displacement  of  the  spot  of  light  on  the  scale 
corresponds  to  any  desired  number  of  amperes  per  centimetre. 
The  second  set  of  strips,  V,  is  put  in  series  with  a  non-inductive 


THE   E.M.F.   OF  AN   ALTERNATOR 


131 


resistance  Es,  which  is  again  adjusted  to  give  the  desired  value  of 
deflection  for  the  pressure  of  supply.  The  motor  is  placed  across  the 
mains  before  the  instrument  as  shown  at  M.  When  this  method  of 
connection  is  used  it  will  be  seen  that  there  is  no  great  difference 
of  pressure  existing  between  any  part  of  the  two  sets  of  strips.  In 


Fig.  103. 

Figs.  93  and  103  are  given  curves  determined  by  the  oscillograph, 
and  it  will  be  noted  that  the  instrument  shows  accurately  even 
the  13th  harmonic  in  one  of  the  cases. 

Example.  The  following  example  will  show  one  of  the  uses  to 
which  the  determination  of  the  instantaneous  curves  of  pressure  and 
current  can  be  applied.  It  refers  to  the  case  of  a  choking  coil  in 
which  the  curves  of  current  and  pressure  (Fig.  104)  were  determined 
by  the  method  first  described.  The  ohmic  resistance  of  the  winding 
on  the  choking  coil  was  so  low  that  the  drop  of  pressure  due  to 
this  cause  was  entirely  negligible,  and  hence  the  induced  E.M.F. 
is  practically  equal  to  the  terminal  pressure.  The  former  is  given 

by  the  relation  e  =  T-^-   where  </>  is  the  flux  in  the  iron  core 

and    T  is   the  number  of  turns  in  the  primary  coil.     We  can 

1    [* 
therefore  write  <£  =  -~  I  edt,  where  <f>  is  the  flux  existing  at  the 

time  t,  hence  the  curve  of  flux  can  be  found  from  the  pressure  curve. 
Two  points  must  be  considered,  first,  it  must  be  remembered  that 
the  whole  length  of  the  abscissa  for  one  period  of  the  curve  is 
equal  to  the  periodic  time  T,  and  hence  any  integration  performed 
must  be  multiplied  by  the  appropriate  factor  to  bring  the  unit  of 

9—2 


132 


ALTERNATING   CURRENTS 


length  of  the  abscissae  into  agreement  with  this  number.  Secondly, 
we  must  see  where  the  flux  curve  is  to  be  reckoned  from ;  since 
the  E.M.F.  is  a  maximum  when  the  flux  is  zero  it  follows  that  we 
must  start  the  integration  of  the  E.M.F.  curve  from  the  point  where 


1-5 


Fig.  104. 

it  has  its  maximum  value.  By  integrating  this  curve  from  that 
point  up  to  any  assumed  set  of  points  the  curve  connecting  the  total 
flux  and  the  time  can  be  found.  If  the  induction  is  required  we 
have  only  to  divide  by  the  area  of  the  iron  core.  The  curve  of 
flux  thus  determined  is  shown  in  the  figure. 

Since  the  simultaneous  values  of  the  flux  and  current  are  thus 
known  it  is  easy  to  plot  the  curve  connecting  the  two  by  simply 
reading  off  the  corresponding  ordinates ;  if  these  be  plotted  against 
one  another  we  shall  evidently  get  the  cyclic  curve  of  the  iron  core. 
When  the  mean  length  of  the  iron  and  the  turns  in  the  primary 
are  known  this  can  evidently  be  expressed  in  terms  of  the  usual 
quantities  B  and  H. 

As  an  example  of  the  reduction  of  such  a  set  of  results  take 
the  curves  just  referred  to.  The  scale  of  time  is  such  that  the 
unit  of  length  corresponds  to  30°,  and  in  this  case  the  periodic 
time  was  ^th  of  a  second,  hence  the  unit  of  length  along  the 

abscissae  is  ^ .  ^  of  a  second.     The  scale  of  pressures  is  such 
oO    \.2t 


THE  E.M.F.   OF  AN   ALTERNATOR  133 

that  the  same  unit  of  length  is  10  volts,  hence  the  area  of  one 

1  108 

square  unit  corresponds  to       —  —      volt-seconds,  or  to 


or  103  lines  of  force,  hence  if  the  volt  curve  be  integrated,  the 
areas  must  be  multiplied  by  this  factor  to  give  the  corresponding 
total  flux.  The  integration  must,  as  has  been  said,  be  started 
from  the  150°  point  where  the  E.M.F.  is  a  maximum,  and  hence 
the  flux  zero.  By  counting  up  the  squares,  starting  from  this 
point,  and  reckoning  up  to  any  other,  and  multiplying  the  result 
by  the  above  factor,  the  flux  existing  at  each  of  the  assigned 
points  was  found,  and  when  plotted  it  gives  the  flux  curve  shown. 
The  cyclic  curve  is  then  readily  obtained  with,  however,  as 
ordinates  the  product  of  flux  and  turns,  and  as  abscissae,  the 
current.  Its  area  a  therefore  gives  the  value  of  /</>  .  T  .  dC.  But 

we    have   <f>  =  Bs  and   H  —  —  •  -=-  ,  where  s  is  the  iron's   cross 

10    i 

section,  and  I  is  the  length  of  the  iron  circuit.  Hence  we  also 
have  a  =  -r—s.llB.dH.  But  we  know  that  -r—lB.dH  is  the 

4-7T  J  47TJ 

energy  required  to  carry  the  flux  round  the  given  cycle,  hence  ^ 

represents  the  energy  in  ergs  for  the  core,  or  a  x  10~8  is  the 
energy  in  joules. 

The  area  must  necessarily  be  interpreted  on  the  scales  of  the 
diagram,  which  are  such  that  unit  length  represents  J  ampere  on 
the  horizontal  scale,  and  2  x  106  lines  of  force  on  the  vertical  one, 
that  is,  each  square  means  106  ergs,  and  in  this  case  the  area  is 
about  42  square  units,  hence  the  energy  required  for  one  cycle  is 
0*42  joule,  or  since  this  work  is  done  in  ^th  of  a  second,  the  rate 
of  loss  of  energy  in  the  core  is  36  watts. 

Since  si  is  the  volume  of  the  iron  it  is  evident  that  by 
dividing  the  result  by  this  volume,  the  loss  per  cubic  centimetre 
per  cycle  can  be  readily  found. 

In  the  case  of  very  large  transformers  the  hysteresis  cycle 
can  be  found  in  the  following  manner,  which  is  based  on  the  same 
considerations  as  the  last.  Let  a  millivoltmeter  be  placed  in  the 
high  tension  side,  and  let  direct  current  be  supplied  to  the  other 
winding  by  means  of  a  battery  and  a  continuously  adjustable 
potential  slide  or  other  device,  which  will  enable  the  current  to 
pass,  without  breaking  the  circuit,  from  a  definite  positive  to  the 
same  definite  negative  value.  If  this  current  is  allowed  in  any 
way  to  change  an  E.M.F.  will  be  produced  in  the  secondary  which 
can  be  read  on  the  millivoltmeter.  Again,  by  a  proper  manipula- 
tion of  the  slide,  this  reading  can  be  kept  at  a  steady  known  value. 
From  a  knowledge  of  the  resistance  of  the  whole  secondary  circuit 
it  is  evident  that  the  rate  of  change  of  flux  corresponding  to 


134  ALTERNATING   CURRENTS 

any  reading  on  the  millivoltmeter  can  readily  be  found,  and  hence 
when  the  reading  is  kept  at  the  desired  steady  value,  the  flux 
existing  in  the  core  of  the  transformer  will  be  found  at  once  from 
the  observation  of  the  time  that  has  elapsed  from  starting  the 
observation.  Hence  simultaneous  observations  of  the  time  and 
current  are  taken  under  the  condition  of  constant  indication  of 
the  millivoltmeter,  the  relation  between  the  total  flux  and  the 
current  flowing  can  be  obtained.  The  area  of  the  cyclic  curve 
thus  obtained  will  evidently  give  the  loss  of  energy  in  hysteresis. 
It  may  be  noted  that  if  by  wattmeter,  or  other  proper  methods, 
the  total  loss  in  the  core  has  been  found,  the  difference  between 
this  amount  and  the  hysteretic  loss  will  be  that  due  to  the  eddy 
currents  in  the  core. 


CHAPTER  X. 

EFFICIENCY  OF   ALTERNATORS. 

The  losses  of  energy.  The  losses  of  energy  in  an  alternator 
fall  into  two  categories,  those  due  to  the  passage  of  the  currents  in 
the  windings  and  those  incident  to  the  rotation.  The  former  are 
two  in  number,  that  involved  in  the  excitation,  Ww,  and  the 
ohmic  loss  in  the  armature,  W0.  Of  these  the  excitation  loss 
is  readily  deduced  from  the  knowledge  of  the  resistance  of  the 
exciting  winding  or  windings  and  the  currents  flowing  therein. 
The  armature  ohmic  loss  can  best  be  found  by  the  method  to 
be  described  later.  The  rotational  losses  are  more  varied.  There 
is  first  the  ordinary  mechanical  loss  due  to  the  friction  of  the 
bearings.  Secondly  there  is  the  hysteretic  loss  in  the  core  of 
the  armature.  Lastly  there  are  various  sources  of  loss  due  to 


Fig.  105. 

the  existence  of  eddy  currents,  which  may  occur  in  different 
ways.  As  in  the  transformer  the  armature  stampings  will  of 
necessity  have  a  certain  amount  of  eddy  currents  induced  in  them 
which  are  reduced  in  amount  as  far  as  is  commercially  necessary 
by  using  stampings  of  the  proper  thinness.  But  eddy  currents 
can  be  produced  in  other  places ;  for  example,  when  the  armature 
is  made  with  teeth  the  induction  they  carry  will  vary  in  density 
along  the  polar  face  (Fig.  105)  and  hence  the  latter  will  experience 
changes  of  flux  as  the  armature  rotates  which  may  have  considerably 
higher  periodicity  than  that  of  the  current.  Such  changes  in 
flux  will  produce  eddy  currents  in  the  poles  themselves  ;  these  can 


136  ALTERNATING    CURRENTS 

be  greatly  reduced  by  the  common  expedient  of  laminating  the 
polar  faces  in  such  a  way  as  to  prevent,  as  far  as  possible,  such 
currents  from  flowing.  Another  place  where  such  eddy  currents 
can  be  produced  is  in  the  substance  of  the  conductors  on  the 
armature.  Thus  from  Fig.  106  it  will  readily  be  seen  that  the 
distribution  of  flux  across  the  conductors,  even  when  they  are 


Fig.  106. 

wound  in  slots,  will  be  different  when  the  slot  is  under  a  pole 
and  when  it  is  leaving  it,  the  consequent  changes  of  induction  will 
again  produce  eddy  currents  in  the  substance  of  the  conductors 
themselves.  The  stray  field  of  the  machine  may  also  cut  metallic 
parts  in  such  a  way  that  currents  can  be  produced.  Hence  the 
rotational  loss  Wr  can  be  considered  as  made  up  of  the  three  parts, 
the  friction  loss,  Wf,  the  hysteretic  loss,  Wh,  and  the  eddy  current 
losses,  We,  so  that  we  can  write  Wr  =  Wf+  Wh  +  We.  Of  these 
three  components  the  first  is  proportional  to  the  speed  only, 
the  second  varies  very  nearly  as  the  speed  and  as  the  1*6 th  power 
of  the  maximum  induction  to  which  the  iron  of  the  core  is 
magnetised,  while  the  third  varies  as  the  square  of  the  speed  and 
the  square  of  the  same  maximum  induction.  In  most  cases  the 
speed  remains  constant  and  hence  the  loss  varies  only  with  the 
induction  in  a  given  machine.  Since  the  excitation  increases 
with  the  load,  the  value  of  Wr  must  increase  also  therewith,  and 
this  effect  will  be  intensified  by  any  distortion  or  other  alteration 
of  the  field  such  as  we  shall  see  will  be  produced  by  the  armature 
current.  The  rejection  of  the  heat  resulting  from  these  losses  is 
carried  out  partly  by  radiation  and  partly  by  the  currents  of  air  pro- 
duced by  the  rotation  of  the  armature.  It  follows  that  for  the  same 
proportional  losses,  the  rise  of  temperature  will  be  considerably 
less  than  in  the  case  of  the  transformer.  A  temperature  test 
must  be  included  in  these  tests  as  in  the  case  of  the  transformer. 

Efficiency.  If  it  were  possible  to  find  the  several  losses 
enumerated  above  when  a  machine  was  delivering  power  denoted 
by  W,  it  is  evident  that  the  input  would  be  given  by 

W+Ww+W0+Wr, 


EFFICIENCY   OF   ALTERNATORS  137 

W 

and  hence  the  efficiency  by  77  =  = — ™ = ^- .     In  many 

rf    ~\~    VVw-\-    W 0  +    Wr 

cases  it  is  quite  impracticable  to  measure  the  input  directly, 
as  the  machines  used  are  of  large  size,  and  even  in  the  case  of 
small  ones  considerable  difficulty  would  be  experienced  in  making 
such  a  determination,  owing  to  the  inaccuracy  of  transmission 
dynamometers,  hence  the  best  way  is  to  determine  in  some 
manner  the  losses  and  thence  deduce  the  efficiency  as  in  the 
case  of  the  transformer.  This  necessitates  the  employment  of 
a  source  of  power  for  such  measurements  from  the  indications  of 
which  the  power  supplied  can  be  readily  found.  One  very  con- 
venient form  is  a  rated  direct  current  motor.  If  a  motor  with 
separate  excitation  be  provided,  and  if  the  losses  in  its  armature 
for  different  desired  speeds  and  currents  have  been  carefully 
determined,  it  is  evident  that  when  observations  of  the  electrical 
input  of  such  a  machine  are  taken  the  actual  nett  power  it  is 
delivering  can  readily  be  deduced.  When  such  a  motor  is  used  to 
drive  the  machine  under  test  by  means  of  a  carefully  prepared 
belt  in  which  the  losses  are  very  small,  we  have  a  ready  means  of 
determining  the  power  delivered  to  that  machine  under  any 
conditions  in  which  it  may  be  working,  up  to  the  full  load  that 
the  rated  motor  can  deliver.  Another  useful  form  of  prime  mover 
consists  in  a  motor  of  any  description  which  is  either  wholly  carried 
on  a  cradle  or  one  in  which  the  field  magnets  are  hung  on  ball 
bearings  on  the  shaft.  In  such  a  case  the  reaction  between  the 
armature  and  the  rest  of  the  motor  produces  a  couple  which  tilts 
up  the  frame  about  the  axis  of  rotation.  By  means  of  weights  the 
original  configuration  can  be  restored,  and  if  these  weights  and 
the  perpendicular  distance  at  which  they  are  hung  from  the 
motor's  shaft  are  known,  it  is  evident  that  this  gives  the  couple 
that  is  being  supplied  to  the  belt.  If  the  energy  loss  incident 
to  bending  the  belt  round  the  pulley  is  negligible,  this  couple 
must  be  equal  to  that  which  the  machine,  under  test,  is  receiving. 
Hence  if  the  speed  of  the  latter  be  measured,  the  power  given  to 
it  is  known.  This  method  is  in  some  respects  better  than  the  use 
of  a  rated  motor. 

No  load  test  and  short  circuit  test.  Such  a  known  prime 
mover  being  available,  let  the  machine  be  excited  to  its  ordinary 
amount  so  as  to  give  the  full  pressure,  the  power  given  will  then 
consist  of  the  value  of  Wr  at  no  load.  As  a  first  approximation 
this  may  be  taken  as  being  very  nearly  constant  over  the  range  of 
operation  of  the  machine  and  hence  this  value  of  Wr  may  be  used 
in  the  determination  of  the  efficiency.  Now  short  circuit  the 
armature  through  a  low  resistance  ammeter  and  adjust  the 
exciting  current  till  any  desired  current  is  flowing  up  to  the 
largest  at  which  the  test  has  to  be  made.  The  losses  supplied  by 
the  rated  motor  are  two  in  number,  that  due  to  the  current 


138 


ALTERNATING   CURRENTS 


passing  in  the  armature,  or  W0,  and  a  certain  amount  of  core  loss. 
But  the  excitation  necessary  to  send  full  load  current  through  the 
armature  on  short  circuit  is  small  compared  with  that  ordinarily 
employed  at  full  pressure,  and  hence  this  test  can  be  taken 
as  approximately  giving  the  value  of  W0  corresponding  to  the 
current  indicated  by  the  ammeter  in  the  armature  circuit.  It 
follows  that  these  two  tests,  together  with  a  knowledge  of  the 
excitation  current  and  the  field  resistance,  give  enough  data  for 
a  close  approximation  to  the  efficiency  of  the  machine  to  be 
found. 

Combined  test.  If  two  similar  machines  are  available  a  test 
similar  to  the  combined  transformer  test  can  be  made.  The  two 
armatures  should  be  rigidly  connected  together,  and  in  so  doing 
it  is  best  to  give  a  small  angle  of  phase  difference  between  the  two. 
The  effect  of  this  is  evidently  to  give  a  resultant  E.M.F.  due  to  the 
two  armatures,  when  equally  excited,  which  is  nearly  in  quadrature 
with  either  (see  Fig.  219),  and  since  the  circuit  of  the  two 
armatures  is  very  inductive,  the  current  will  again  lag  nearly  90° 
behind  this  current  and  will  thus  be  roughly  in  phase  with  the 
pressure  due  to  either.  The  circulating  power  is  measured  by 
a  wattmeter,  W,  as  shown  in  Fig.  107,  and  this  condition  is,  as  we 
have  seen,  favourable  to  the  accuracy  of  indication  of  the  same. 


Fig.  107. 

The  initial  phase  angle  must  be  fairly  small  to  prevent  the  mini- 
mum difference  of  E.M.F.  which  occurs  for  equality  of  the  machines' 
E.M.F.s  being  too  large.  By  regulating  the  excitation  of  one  of  the 
machines  the  load  can  be  adjusted  to  the  desired  full  amount. 
The  two  coupled  machines  are  driven  as  in  the  last  case  by  the 
rated  motor  from  which  the  loss  of  power  is  found.  Let  the 
power  circulating  between  the  machines  be  W  as  shown  by  the 
wattmeter  and  the  power  lost  be  WL,  as  shown  by  the  ammeter 
and  voltmeter  V  and  A  after  deduction  of  the  internal  loss  on  the 
motor,  and  let  it  further  be  assumed  that  half  the  lost  power  goes  to 


EFFICIENCY   OF   ALTERNATORS  139 

each  machine,  and  that  their  efficiencies,  77,  are  the  same,  which 
will  be  very  nearly  true  for  machines  of  fair  size.     The  generator 

W  (  W  \ 

then  absorbs  W  +  —^  watts  and  therefore  produces  77  (  W  H  —  ^  ) 

W 
watts.      The  motor   delivers    W  --  ~    watts    and   consequently 

absorbs    - 


which  leads  to 


This  test  gives  more  nearly  the  losses  that  are  incident  to  full  load 
conditions  than  does  the  last  one. 

It  would  of  course  be  possible  to  make  the  test  even  more 
similar  to  the  ordinary  direct  current  one  by  supplying  the  power 
by  means  of  alternating  currents,  in  which  case  the  machines 
would  act  as  "  synchronous  "  motors,  and  the  lost  power,  WL,  would 
be  measured  directly  by  a  wattmeter  in  the  supply  circuit. 

Deceleration  tests.  The  following  method  enables  the  actual 
losses  to  be  found  and  the  separation  between  the  hysteretic  and 
eddy  current  parts  of  the  no  load  loss  to  be  effected,  the  frictional 
couple  having  been  previously  found.  Let  the  armature  of  the 
machine  be  permitted  to  slow  down  to  rest  from  its  normal  speed ; 
if  P  denote  the  retarding  torque  at  any  instant  due  to  any  loss 
that  is  occurring  in  the  armature,  and  if  &>  be  its  angular  velocity 

at  that  instant  we  have  the  relation  P  =  I0 .  -j-  ,  where  /„  is  the 

at 

moment  of  inertia  of  the  armature,  etc.  If  W  denote  the  rate  of 
working  at  that  instant  we  further  have 

j-        day 
°'CO~dt' 

Let  the  curve  in  Fig.  108  give  the  relation  between  the  speed 
or  angular  velocity  and  time  for  the  machine  slowing  to  rest  under 
the  given  circumstances,  then  if  the  normal  at  any  point  be  drawn 
and  if  N  denote  the  length  of  the  subnormal  MN  at  that  point 

P  we  have  also  N=  o>  -7-  ,  and  thus  we  can  write  W  =  I0 .  N.     It 

follows  that  the  loss  at  any  speed  can  be  found  if  the  above  curve 
and  the  value  of  I0  are  found,  by  drawing  the  subnormal  at  the 
points  corresponding  to  the  required  speeds.  The  curve  can  be 
determined  as  follows.  Let  the  machine  be  driven  by  a  motor  as 


140 


ALTERNATING   CURRENTS 


before,  and  at  a  given  moment  let  the  driving  power  be  cut  off,  the 
excitation  being  kept  constant.  Take  the  time  occupied  in  falling 
to  any  known  speed  as  shown  by  a  tachometer  driven  by  the 
machine.  If  this  is  repeated  for  many  different  final  speeds  down 
to  the  time  taken  to  come  to  rest,  the  curve  required  can  readily 
be  found.  It  remains  to  measure  the  value  of  the  constant  I0.  In 


N     Time 

Fig.  108. 

general  the  form  of  the  armature  is  too  complex  to  allow  this 
being  calculated  from  the  drawings,  but  it  can  be  found  as  follows. 
Let  the  rated  motor  be  employed  to  determine  accurately  the 
power  W±  required  to  drive  the  machine  steadily  at  the  given 
excitation,  and  let  the  subnormal  at  that  point  be  measured  and 
have  the  length  N^.  Then  evidently  we  have  Wl  —  I0N1,  which 
gives  the  value  of  the  constant  I0. 

Since  the  excitation  is  fixed  the  total  loss  in  this  case  can  be 
written  in  the  form  W  =  a&>  +  6&>2,  where  a  includes  the  friction 
and  hysteretic  loss  and  b  the  eddy  current  one,  and  hence  a  set  of 
observations  of  the  corresponding  speed,  o>,  and  loss,  W,  will 
enable  the  values  of  a  and  6  to  be  found,  and  the  separation  of 
the  two  categories  of  loss  can  be  carried  out.  For  since  the 
corresponding  values  of  W  and  o>  have  been  found,  for  each  value 

,  W 
of  the  latter  we  can  determine  the  value  of  —  ,  and  if  a  curve  be 

W 
drawn  connecting  the  quantity  —  with  w  it  will  have  the  equation 

W 

—  =  a  +  bo) ;    thus  the  intercept  on  the  axis  gives  the  value  of 

a  and  the  slope  of  the  curve  the  value  of  b. 

The  following  modification  of  the  method  (due  to  Dr  Sumpner) 

enables  the  value  of  -y-  to  be  found  with  great  accuracy  and  hence 
(lit 

does  not  involve  the  determination  of  the  full  deceleration  curve. 
It  requires  the  use  of  the  driving  motor  (which  in  cases  where  an 


EFFICIENCY   OF   ALTERNATORS 


141 


exciter  is  attached  to  the  dynamo,  may  be  that  machine),  as 
an  indirect  method  of  measuring  the  speed.  Let  the  motor  be 
constantly  excited  as  shown  in  Fig.  109 ;  then  if  E  is  the  E.M.F.  it  is 
producing,  this  E.M.F.  will  be  proportional  to  the  speed  or  we  shall 


Fig.  109. 

have  the  relation  b .  E  =  co.  Let  the  pressure  between  the  mains 
that  are  providing  the  motive  power  be  V,  and  let  a  voltmeter  of 
low  range  be  placed  across  the  terminals  of  a  switch  by  which  the 
current  can  be  cut  off  from  the  motor's  armature.  If  v  be  the 
reading  of  this  instrument  we  evidently  have  v  =  V  —  E  and  thus 

dv     dE      ,  .  ,   .      ,        da)     ,    dv       0. 

we  also  have  -j-  =  -7- ,  which  leads  to  -j-  =  6  .  -j-  .     Since  the  volt- 
at      at  at          at 

meter  is  one  of  short  range  a  large  deflection  will  be  produced  by 
but  a  small  change  of  speed.  When  the  full  deflection  has  been 
attained  the  switch  must  be  at  once  closed  to  speed  up  the  machine 
again.  Hence  if  the  normal  speed  and  pressure  at  no  load  of 
the  motor  are  found,  this  gives  the  value  of  6,  and  if  in  addition 
the  time  dt  taken  for  the  small  alteration  of  pressure  dv  be 

measured,  we  have  a  close  approximation  to  the  value  of  6 .  -=-  and 

hence  to  that  of  -=- .     In  what  follows  it  will  be  assumed  that  the 
dt 

angular  acceleration  or  retardation  is  measured  in  this  manner. 

The  torque  at  normal  excitation  can  be  found  as  follows  with 
the  above  method  of  measurement.     Let  the  deceleration  under 

these  conditions  be  found  and  let  it  be  denoted  by  (-?: )  >  we  have 

the  relation  P  =  I0  (  -=- ) .     Then  by  means  of  a  brake  put  over  the 
V  rf*/i 

pulley  of  the  machine,  or  in  any  other  suitable  manner,  let  an 
additional  retarding  couple  of  the  known  amount  P0  be  applied  and 
again  determine  the  angular  deceleration.  We  then  have 

-  T  (dw\ 

-J 


142  ALTERNATING   CURRENTS 

Hence  the  value  of  I0  can  be  eliminated  and  that  of  P  found,  from 
which  the  loss  at  normal  speed  can  at  once  be  determined.  In 
a  certain  small  machine  it  was  found  that  at  660  R.P.M.  the  driving 
motor,  which  was  directly  attached,  gave  an  E.M.F.  of  100  volts. 
Hence  the  value  of  u>  is  69'5  and  that  of  b  is  0*695.  The  mean  of 
several  tests  gave  3  seconds  as  the  time  for  the  auxiliary  voltmeter 

to  read  10  volts,  hence  the  value  of  -j-  is  -=-  and  that  of  (  -y-  )   is 

at       o  \dt/i 

2'32.  A  band  brake  giving  a  torque  of  3'26  foot-pound  units  was 
put  on  the  machine  and  then  it  was  found  that  it  took  1'6  seconds 

for  the  same  change  of  pressure,  giving  as  the  value  of  (  -j-  )    the 

\dt  /2 

amount  4'33.  We  then  have  P  =  2'32  10  and  P  +  3'26  =  4'33  .  10 
which  leads  to  P  =  3-l76  foot-pound  units.  This  is  nearly  the  value 
of  the  no  load  torque,  and  since  the  normal  speed  was  11  R.P.S.  the 
rate  of  loss  of  energy  is  3*76  X  69*5  =  260  foot-pounds  per  second  or 
352  watts. 

The  application  of  the  extra  retarding  torque  can  be  made  by 
permitting  the  machine  to  supply  a  current  to  a  non-inductive 
resistance.  For  the  purpose  of  this  method  it  is  best  to  slightly 
modify  the  expression  used.  Let  £0  denote  the  normal  pressure 
produced  by  the  alternate  current  armature,  at  constant  excitation 
this  will  be  nearly  proportional  to  the  speed,  and  we  can  conse- 

quently  write    aS0  =  &>.      But    we    have   P  =  I0  -=-    and  hence 

cut 

PCD  =  70ft)  -j-  .     But  as  before  Pay  is  the  loss  of  power  that  has  to 

be  found.  It  can  be  considered  as  equivalent  to  a  current  <@0 
which  is  in  phase  with  the  pressure  of  the  machine,  and  hence 
we  have 

da> 


or  on  substituting  for  &>  we  get 


The  value  of  -7-    can  be   found   as  in  the   last   case.     First   let 
dt 

the  machine  be  allowed  to  drop  in  speed  the  desired  amount  when 
the  armature  is  on  open  circuit  and  let  the  deceleration  be  found 
as  before,  we  have 


Again,  let  it  decelerate,  but  let  it  be  delivering  a  known  current 
to  a  non-inductive  resistance,  we  must  then  have 


EFFICIENCY   OF   ALTERNATORS  143 


From  these  equations  the  quantity  al  can  be  eliminated  and  hence 
the  value  of  ^  0  found,  which  immediately  gives  the  no  load  loss  in 
the  form  &0<&0. 

The  accompanying  core  loss  in  the  motor  has  been  neglected 
in  each  case  ;  in  general  it  would  be  small  compared  with  that  in 
the  machine,  but  if  necessary  the  proper  correction  can  readily  be 
applied. 


CHAPTER  XL 

POLYPHASE  E.M.F.S  AND  CURRENTS. 

Two-phase  dynamos.  In  the  monophase  dynamo  we  saw 
that  it  was  not  usual  to  utilize  all  the  available  space  on  the 
armature  core  for  winding  coils,  and  thus  it  is  possible  to  place  a 
second  set,  or  even  two  other  sets,  of  coils  on  the  same  armature 
core.  In  the  former  case  the  second  set  of  coils  could  be  wound 
with  their  centres  midway  between  the  original  set  as  shown  in 
Fig.  110,  where  the  coils  marked  A  are  the  original  ones  and  those 


Fig.  110. 

marked  B  are  the  new  ones.  It  is  evident  that  in  such  a  case  the 
E.M.F.  generated  in  the  set  of  coils  A  will  be  so  related  in  time  to 
that  in  the  set  B  that  the  maximum  E.M.F.  in  A  is  produced  at 
the  instant  the  E.M.F.  in  B  is  zero,  or  the  E.M.F.S  in  the  two  armatures 
are  in  quadrature  as  regards  phase ;  such  a  machine  is  said  to  be 
a  two-phase  dynamo. 

As  in  the  ordinary  alternator  we  may  have  a  winding  with  as 
many  coils  in  each  set  as  there  are  poles,  such  as  the  one  in 
Fig.  110,  or  we  may  have  a  hemitropic  form  with  fewer  coils 
than  poles,  as  shown  in  Fig.  Ill ;  certain  relations  must  be 


H     [  J  —  1>  ^  —  },  £- 

-^H 

Fig.  111. 

fulfilled   in    such    cases,    for    details    of    which    the   student   is 
referred   to  the  larger  books.     Each  set  of  coils  can  be  wound 


POLYPHASE  E.M.F.S  AND  CURRENTS 


145 


in  various  ways  with  concentrated  or  distributed  windings 
and  with  series  or  parallel  arrangements  as  in  the  monophase 
machine.  Fig.  110  would  represent  a  simple  case  of  series  arrange- 
ment, and  in  this  case  the  two  armatures  are  quite  distinct  from 
one  another,  and  the  two  ends  of  each  set  of  windings  are 
attached  to  a  pair  of  slip  rings  in  the  same  manner  as  the  single 
winding  of  the  monophase  dynamo.  In  Fig.  112  is  shown  a  form 
of  completely  continuous  winding,  which  here  takes  the  form  of 


Fig.  112. 

an  ordinary  Gramme  ring:  if  collecting  points  be  fixed  at  distances 
apart  equal  to  half  the  pitch  of  the  poles,  and  if  the  alternate 
points  be  attached  to  slip  rings,  it  is  evident  that  the  winding 
forms  a  completely  distributed  one  with  parallel  arrangement  of 
the  circuits,  there  being  as  many  parallel  circuits  as  there  are 
pairs  of  poles.  If  points  midway  between  the  first  set  of  points  be 
similarly  joined  to  two  rings,  they  will  give  a  second  winding  in 
which  the  flux  is  zero  when  the  flux  through  the  first  set  is  a 
maximum,  and  will  thus  form  an  armature  in  which  the  E.M.F.  is  in 
quadrature  with  that  in  the  first  one.  In  this  case  it  is  evident 
that  the  armatures  cannot  be  treated  as  independent.  Such  a 
form  of  winding  will  be  obtained  if  the  armature  of  an  ordinary 
direct  current  multipolar  dynamo  has  the  appropriate  points 
joined  up  to  four  rings.  A  machine  of  this  form  is  called  a  Rotary 
Converter,  and  will  be  treated  of  more  fully  later  on.  If  the 
direct  current  winding  be  not  a  simple  Gramme  winding  but  some 
one  of  the  many  forms  of  drum  windings,  it  is  still  possible  to 
find  points  in  the  armature  that  very  approximately  fulfil  the 
required  conditions ;  the  arrangement  of  such  windings  is  beyond 
the  intended  scope  of  this  book ;  the  student  is  referred  to  Prof. 
S.  P.  Thompson's  work  on  Polyphase  Currents  for  full  details. 

Vector  representation.  We  must  now  see  how  the  pressures 
and  currents  in  these  cases  can  be  represented  by  vectors.  It  is 
usual  for  the  E.M.F.S  produced  by  the  two  armatures  to  have  the 
same  virtual  value,  and  we  will  take  this  to  be  the  case.  Draw 
the  vector  OA,  Fig.  113,  to  represent  the  maximum  E.M.F.  in  A's 
armature,  then  if  the  E.M.F.  be  simple  harmonic  the  projection  of 
this  line  on  any  line  rotating  at  a  number  of  revolutions  per 

L.  10 


146  ALTERNATING  CURRENTS 

second  equal  to  the  periodicity  will  represent  the  instantaneous 
E.M.F.  of  the  armature  A.  Similarly,  if  0^  be  any  equal  line  at 
right  angles  to  OA  its  projections  on  the  same  line  will  give  the 
corresponding  value  of  the  E.M.F.  of  B.  In  the  case  where  the  two 
armatures  are  quite  independent  there  is  no  connection  between 


0 

0, > B 

Fig.  113.  Fig.  114. 

A  and  B,  and  hence  we  cannot  say  what  is  the  pressure  between 
the  other  ends  of  A  and  B,  that  is  between  0  and  0'  or  A  and  B. 
Now  let  one  end  of  each  set  of  coils  be  connected.  Then  the  points 
0  and  Ol  become  the  same  point  and  Fig.  114  gives  the  vector 
representation  of  this  case.  The  question  arises,  What  is  the 
vector  representing  the  pressure  between  the  free  ends  of  A  and  B  ? 
It  should  be  noted  that  the  pressure  between  A  and  B  is  not  the 
sum  of  the  E.M.F.s  in  the  armatures  but  the  difference,  and  thus 
the  vector  representing  it  will  be  found  by  reversing  one  vector 
and  combining  this  with  the  other,  or  more  simply  and  generally  by 
joining  A  to  B.  The  direction  in  which  this  vector  is  to  be 
reckoned  depends  on  which  of  the  points  A  or  B  is  taken  as  the 
point  of  reference.  Hence  with  sinusoidal  E.M.F.S  the  pressure 
between  the  free  ends  of  the  coupled  armatures  will  be  \/2  times 
the  E.M.F.  in  either,  and  under  these  circumstances  the  same 
relation  will  apply  to  the  virtual  pressure  between  the  mains 
joined  respectively  to  the  points  corresponding  to  A,  B  and  0. 

Balance.  If  the  two  armatures  be  delivering  current  to 
given  circuits  it  may  be  the  case  that  the  currents  and  phase 
angles  are  different  for  the  two.  We  will  assume  that  this  is  not 
so,  but  that  the  currents  in  the  two  circuits  and  the  phase  angles 
between  the  pressures  and  the  currents  are  the  same  for  both,  in 
which  case  it  is  said  that  the  machine  is  working  on  a  balanced 
load.  Let  XD,  Fig.  115,  be  the  vector  for  the  current  from  As 
armature  and  XE  that  for  the  B's.  Since  the  two  currents  have 
the  same  lead  or  lag,  X,  on  their  respective  pressures  these  two 
vectors  are  also  at  right  angles.  With  the  common  junction 
existing  it  is  evident  that  the  current  flowing  through  the  main 
attached  to  it  will  be  equal  to  the  sum  of  the  currents  in  the  other 
two  mains,  or  will  be  given  by  the  vector  XU.  Its  value  with 


POLYPHASE  E.M.F.S  AND  CURRENTS 


147 


sinusoidal  currents  will  be  \/2  times  either  of  the  components,  and 
its  phase  angle  with  the  pressure  between  the  ends  of  the  outside 
mains  will  be  (90°+X).  When  the  load  is  non-inductive,  the 
current  in  the  common  main  is  in  quadrature  with  the  pressure 
existing  between  the  outside  mains. 

Neutral  point.  As  an  example  of  another  possible  arrange- 
ment of  the  vectors  representing  the  E.M.F.S  take  the  form  of 
winding  shown  in  Fig.  112.  The  vectors  giving  respectively 
the  potential  differences  between  the  two  armatures  must  be 

a 


Bi 

Fig.  116. 

of  the  same  length  and  at  right  angles,  as  in  the  last  case ; 
let  them  be  represented  by  AAl  and  BBl  in  Fig.  116.  Here 
there  is  of  necessity  another  condition  which  must  be  fulfilled, 
namely  that  the  four  potential  differences  between  AB1}  B1Al) 
A1B  and  BA  must  from  symmetry  be  all  equal  in  amount. 
It  follows  that  the  relative  position  of  the  vectors  for  AA^  and 
BBl  must  be  such  that  they  cross  at  right  angles  at  the  centre, 
so  that  the  complete  representation  of  this  case  is  the  square 
shown.  In  this  case  if  the  potential  difference  across  any 
opposite  pairs  of  mains  connected  to  the  rings  is  £,  that  across 
any  adjacent  pair  is  <f/\/2.  The  centre  of  the  square  is  a  point  of 
symmetry  and  is  called  the  neutral  point  of  the  system  of  vectors, 
the  corresponding  point  in  the  armature,  whether  really  existent  or 
not,  being  the  neutral  point  in  the  armature's  winding.  The 
position  of  such  a  point  is  sometimes  a  very  important  matter  to 
bear  in  mind.  In  certain  forms  of  distributed  windings,  such  as 
those  derived  from  direct  current  drum  windings,  the  number  of 
sections  in  the  winding  often  does  not  permit  of  division  into  four 
exactly  equal  parts.  In  such  a  case  the  vector  square  will  be 
slightly  deformed,  and  its  diagonals  will  no  longer  be  at  right 
angles. 

Three-phase  dynamo.  In  the  case  where  the  space  on 
the  armature  is  used  to  wind  three  sets  of  wires  instead  of  two, 
the  sets  of  coils  are  in  general  so  arranged  that  the  phase  difference 

10—2 


148 


ALTERNATING   CURRENTS 


between  the  three  equal  E.M.F.s  produced  is  120°,  that  is,  if  one  be 
represented  by  e—  E  sinpt  the  other  two  will  be  given  by 

e  =  E  sin  (pt  +  JTT)  and  e  =  E  sin  (pt  +  |TT). 

This  will  result  in  the  trace  of  the  E.M.F.S,  when  sinusoidal,  being 
as  shown  in  Fig.  11*7.     Take  the  case  of  an  armature  winding  as 


Fig.  117. 

shown  in  Fig.  118.  The  distance  from  one  north  pole  to  the  next 
corresponds  to  360  electrical  degrees,  hence  if  two  other  sets  of 
coils  B  and  C  are  wound  as  shown  in  addition  to  the  original 


Fig.  118. 

set,  A,  the  E.M.F.s  generated  in  them  will  differ  by  the  required 
angular  amount.  The  corresponding  case  with  the  full  number  of 
armature  coils  is  given  in  Fig.  119.  Just  as  before,  the  several  coils 
can  be  wound  in  either  concentrated  or  distributed  manner  and 


Fig.  119. 

in  series  or  parallel.  The  phase  relation  will  be  fulfilled  in  the  case 
of  a  completely  distributed  winding,  such  as  that  of  a  direct 
current  armature,  by  supplying  rings  attached  to  three  points  at 
120°  instead  of  the  four  points  and  rings  of  the  two-phase  case. 


POLYPHASE  E.M.F.S   AND   CURRENTS 


149 


The  representation  of  this  case  when  the  armatures  are  quite 
distinct  will  be  by  means  of  three  equal  vectors  at  120°  as  shown 
in  Fig.  120  at  A,  B  and  C. 

Star  connection.  There  are  two  methods  open  to  us  for 
diminishing  the  mains  required.  If  the  terminals  of  armature  A 
be  called,  as  shown  in  Fig.  121,  1  and  2,  those  of  B,  3  and  4,  and 
those  of  C,  5  and  6,  we  can  connect  up  1,  3  and  5  into  a  common 

<{* 


Fig.  120. 


Fig.  121. 


point,  and  in  this  case  the  connection  is  called  the  T  or  star 
connection.  The  vector  representation  of  the  pressures  will  then 
be  as  at  Fig.  122,  where  the  three  vectors  are  OA,  OB  and  00.  As 
in  the  two-phase  case,  the  vectors  giving  the  pressures  existing 
between  adjacent  mains  will  be  AB,  BC  and  CA,  and  for  the  case 
where  the  initial  E.M.F.S  are  sinusoidal,  each  of  these  is  evidently 
V3  times  the  pressure  between  the  armature  terminals ;  0  is  the 
neutral  point  of  the  three  armatures.  If  we  take  the  external 


Fig.  122. 


B 


Fig.  123. 


Fig.  124. 


circuits  as  non-inductive  and  equally  loaded,  the  vectors  representing 
the  three  currents  will  necessarily  be  parallel  to  OA,  OB  and  00, 
and,  since  there  can  be  no  current  flowing  to  or  from  the  neutral 
point  0,  these  vectors  must  be  arranged  in  a  triangle  as  at  Fig.  123, 
with  the  sides  parallel  to  the  above  vectors.  It  follows  that  in 


150 


ALTERNATING   CURRENTS 


this  case  the  pressure  between  any  two  mains  is  at  right  angles  to 
the  current  in  the  opposite  main.  For  inductive  loads  the  current 
triangle  must  be  turned  in  the  proper  direction  through  the  angle 
of  lead  or  lag,  X,  as  shown  in  Fig.  124,  the  phase  difference  between 
any  main  current  and  the  pressure  between  the  opposite  pair  of 
mains  is  in  that  case  (90°  +  X) :  it  may  also  be  noted  that  the 
current  in  any  main,  such  as  A,  Fig.  121,  has  the  phase  angle 
(30° +X)  relative  to  the  pressure  between  that  main  and  one 
adjacent  main,  and  the  phase  angle  (30°  —  X)  relative  to  the 
pressure  between  that  main  and  the  other  adjacent  one. 

Mesh  connection.  Another  method  of  combination  of  the 
six  ends  of  the  armatures  would  be  to  take  them  two  and  two  in 
pairs ;  this  is  called  the  A  or  mesh  connection.  In  this  the 
vector  representation  of  the  pressures  is  simply  a  triangle  as  in 
Fig.  125.  As  regards  the  currents  it  must  be  borne  in  mind  that 
if  we  call  a,  b  and  c  the  currents  in  the  three  armatures  and 
denote  those  in  the  mains  by  ab,  be  and  ca,  the  current  ab  must 
not  be  looked  on  as  the  resultant  of  a  and  b  but  as  being  that 


Fig.  125. 


Fig.  126. 


current  which,  when  combined  with  a,  will  leave  b ;  that  is,  b  is 
the  resultant  of  a  and  ab  and  vice  versa.  Thus  if  the  triangle  PRQ 
(Fig.  126)  be  drawn  for  the  case  where  the  load  is  non-inductive, 
with  its  sides  each  equal  to  the  armature  currents  and  parallel  to 
the  pressure  vectors,  and  if  the  three  lines  PS,  QT  and  EU  be 
drawn  bisecting  the  angles  of  this  triangle  and  of  length  \/3  times 
the  length  of  those  sides,  it  will  be  seen  from  the  construction 


POLYPHASE  E.M.F.S  AND  CURRENTS  151 

that  these  lines  must  be  the  vectors  representing  the  several  line 
currents.  Thus  as  before  the  resultant  current  in  a  main  is,  with 
non-inductive  loads,  at  right  angles  to  the  pressure  between  the 
opposite  mains,  but  the  current  in  the  mains  is  \/3  times  that 
in  the  armatures,  the  pressures  between  the  mains  being  the 
same  as  those  produced  by  the  armatures.  With  a  phase  angle, 
X,  the  vector  figure  of  the  currents  must  be  turned  through 
that  angle,  as  in  the  star  case.  The  above  numerical  re- 
lations are  evidently  only  true  for  the  case  where  the  pressures 
and  currents  are  simple  harmonic  quantities,  and  for  other  forms 
of  curves  these  relations  will  not  necessarily  hold  good.  For 
example,  if  in  the  mesh  connection  the  third  harmonic  be  present 
in  the  current  curve  it  is  evident  that  since  the  three  curves 
differ  in  phase  as  regards  their  fundamentals  by  120°,  this  harmonic 
will  just  be  in  phase  in  each  of  the  armatures  and  will  hence 
merely  cause  a  current  to  circulate  locally  round  the  mesh.  Such 
complex  curves  cannot  be  properly  represented  by  the  relations  we 
have  derived.  In  practice,  however,  the  difference  is  not  great 
between  the  results  of  a  vectorial  treatment  and  the  results  of 
experiment. 


CHAPTER  XII. 

MEASUREMENT  OF   POLYPHASE   POWER. 

Balanced  loads.  We  must  now  consider  the  question  of 
power  measurements  in  a  polyphase  system.  With  two  phases 
and  a  balanced  load  it  is  only  necessary  to  connect  a  wattmeter  in 
one  of  the  circuits  in  the  ordinary  way  and  twice  the  reading  will 
give  the  power  transmitted.  If  the  circuits  are  unbalanced  two 
wattmeters  would  be  required,  one  in  each  circuit.  These  can  be 
mechanically  connected,  that  is  to  say,  the  two  shunt  coils  can  be 
fixed  to  the  same  spindle  and  pointer,  and  if  the  two  instruments 
have  been  so  arranged  that  their  calibrations  are  the  same,  the 
total  power  can  be  read  at  one  observation.  In  the  case  of  a 
three-phase  balanced  load,  if  £  be  the  pressure  at  the  terminals  of 
any  one  of  the  three  armatures  and  if  Vj>  be  the  current  it  is 
delivering,  the  total  power  delivered  will  be  W  =  3 .  S^  cos  X, 
where  X  is  the  phase  angle  between  the  pressure  and  current. 


1  0' 


Fig.  127. 

Now  it  is  not  always  possible  to  connect  up  a  wattmeter  with  the 
shunt  coil  across  one  armature  and  the  series  coil  in  the  armature 
circuit,  the  series  coil  must  be  in  one  of  the  mains  and  the 
shunt  one  across  a  pair  of  mains,  since  the  neutral  point  is  not 
usually  accessible.  The  difficulty  can  be  overcome  by  the  use 
of  a  neutral  point  resistance  as  shown  in  Fig.  127.  Three  equal 
non-inductive  resistances  are  arranged  as  a  star  across  the  supply 
mains,  and  the  shunt  circuit  is  connected  to  the  common  point,  0le 


"MEASUREMENT  OF  POLYPHASE  POWER  153 

The  whole  resistance  of  the  shunt,  the  series  resistance  and  one 
of  the  branches  of  the  auxiliary  star  must  be  made  of  the  right 
amount  for  the  wattmeter's  shunt  circuit.  In  this  case  with 
balanced  load  it  is  evident,  from  symmetry,  that  the  points  0  and 
Ol  must  be  at  the  same  pressure  and  thus  the  wattmeter  will 
read  one-third  of  the  output  of  the  dynamo. 

A  method  based  on  the  assumption  of  sine  variation  of  currents 
or  pressures,  that  can  be  applied  to  the  case  of  balanced  loads,  is 
the  following.  Consider  the  mesh  connection  for  a  balanced  load, 
the  pressure  between  the  mains  is  due  to  one  of  the  armatures 
and  is  therefore  £,  while  the  current  in  the  main  has  a  value  V3 
times  that  in  either  armature.  Again,  a  reference  to  p.  150  will 
show  that  the  angles  between  the  current  in  any  main  and  the 
pressures  between  that  and  the  adjacent  mains  are  respectively 
(30°  -  X)  and  (30°  +  X).  Let  a  wattmeter  (Fig.  128)  be  connected 
with  its  series  coil  in  one  main,  and  let  the  shunt  be  first 


O 


Q 
Fig.  128. 

connected  to  the  point  P  and  then  to  the  point  Q.  In  the  first 
case  its  reading  will  measure  the  quantity  V3 .  £*$ .  cos  (30°  —  X) 
and  in  the  second  the  quantity  \/3 .  S .  *$ .  cos  (30°  4-  X).  The  sum 
of  its  readings  will  then  give  V3  .  g<@ .  (cos  (30°  +  X)  +  cos(30°  -  X)} 
which  reduces  to  V3 .  S.  <@ .  (2  cos  30° .  cos  X)  or  3  .  g<@ .  cos  X,  that 
is  to  say,  the  power  that  the  load  is  taking.  The  addition  of  the 
two  readings  can  be  made  automatically  if  the  points  P  and  Q 
are  both  joined  to  the  shunt  of  the  wattmeter  by  means  of 
equal  high  resistances,  but  the  constant*  of  the  instrument  must 
be  taken  with  only  one  of  them  in  circuit.  It  is  evident  that 
exactly  similar  considerations  will  apply  to  a  load  arranged  in  a 
star  fashion.  It  must  be  remembered  that  this  method  not  only 
involves  the  assumption  that  the  loads  are  balanced,  but  also  that 
the  pressures  or  currents  vary  as  sines. 

The  following  method  of  expressing  the  power  in  a  three- 
phase  system  is  sometimes  employed.  Consider  the  case  of  a 
star  connection  with  balanced  load,  and  let  the  power  have  been 
measured  as  described.  If  W  denote  this  power  and  £  and  ^  the 
currents  and  pressure  due  to  each  of  the  similarly  loaded  armatures, 
the  total  power  will  be  given  by  the  relation  W  —  3 .  S^ .  cos  X. 


154 


ALTERNATING   CURRENTS 


Now  let  Sm  denote  the  pressure  existing  across  any  two  of  the 
mains,  which  is  in  general  the  only  pressure  that  can  be  readily 
measured,  then  on  the  assumption  of  sinusoidal  pressures  we 
know  that  the  numerical  value  of  this  virtual  pressure  is  \/3  times 
that  of  the  pressure  £  contributed  by  each  armature,  hence  we 
can  write  £m  =  *J%.£  or  W=  \/3.  ^.^.cosX.  In  using  this  ex- 
pression, however,  care  must  be  exercised.  It  does  not  denote 
that  the  power  has  been  measured  by  joining  up  a  wattmeter 
with  its  series  coil  in  one  main  of  the  star  and  the  shunt 
across  the  adjacent  one,  all  it  denotes  is  that  we  have  separately 
measured  the  power  taken  by  the  whole  apparatus,  the  current  in 
the  main,  and  the  pressure  between  two  mains,  and  for  convenience 
write  the  relation  between  the  three  in  this  way. 

Similarly  in  the  case  of  the  mesh  connection  with  balanced 
power,  if  £,  <$  and  W  have  the  same  meanings  as  before,  we 
can  measure  W  and  £  but  not  ^;  all  we  can  do  is  to  measure 
the  current  in  one  of  the  mains  attached  to  a  junction  of  the  rnesh. 
If  this  be  called  ^m,  on  the  sinusoidal  assumption  we  again  have 
<^  =  V3  <$m  and  thus  arrive  at  W  =  V3 .  &@ .  cos  X,  where  the  three 
quantities  W,  £  and  <$  are  directly  measured.  The  same  point 
arises  as  before,  that  is,  the  current  ^  must  be  taken  to  denote 
solely  its  virtual  value  and  it  must  not  be  taken  to  connote  its 
phase  relationships. 

Unbalanced  load.  It  can  be  shown  as  follows  that  by 
means  of  two  wattmeters  we  can  measure  the  power  of  a  three- 
phase  dynamo  whether  the  load  be  balanced  or  not.  Consider 


Fig.  129. 

the  star  winding  shown  in  Fig.  129  and  let  the  two  wattmeters  be 
connected  as  indicated.  Let  ea,  eb  and  ec  be  the  pressures  existing 
at  any  instant  between  the  neutral  point  and  the  ends  of  the  three 
armatures,  and  let  ca,  cb  and  cc  be  the  corresponding  currents  in 
those  armatures.  The  instantaneous  power  being  delivered  will  be 


MEASUREMENT  OF  POLYPHASE   POWER  155 

and  the  mean  power  over  the  period  will  be 

1  CT 

W=-  I    (eaca  +  ebcb  +  eccc)dt 


where  r  is  the  periodic  time.     But  in  the  star  case  we  have 

ca  +  Cb  +  cc  =  0, 

and  thus  ec  (ca  +  cb  +  cc)  =  0. 

Hence  by  subtracting  this  expression  from  that  under  the  integral 

1  fT 

we  get  W  =  —  /    ca  (ea  —  ec)  +  cb  (eb  —  ec)  dt. 

T  Jo 

But  from  the  method  in  which  the  two  wattmeters  are  connected 
it  is  seen  that  the  right  hand  of  this  expression  is  what  the  two 
instruments  measure,  and  thus  two  wattmeters  connected  as 
shown  will  measure  the  power  under  any  conditions  of  the  three 
circuits.  The  mesh  case  is  left  to  the  student  to  prove.  It  will 
be  seen  that  for  some  conditions  of  phase  angle  in  the  circuits  one 
of  the  wattmeters  may  register  negative  power;  to  avoid  any 
difficulty  it  is  desirable  to  combine  the  two  spindles  mechanically 
as  mentioned  in  the  two-phase  case. 

Constancy  of  output  with  balanced  load.     It  may  be 

noted  that  the  polyphase  dynamo  working  on  a  balanced  load  has 
one  advantage  over  the  monophase  one  in  the  constancy  of  the 
rate  of  production  of  energy.  In  the  latter,  even  in  the  case  of 
unit  power  factor,  the  delivery  of  power  falls  to  zero  twice  per 
alternation,  and  with  a  phase  angle  is  negative  for  two  portions  of 
each  alternation.  Take  the  case  of  a  two-phase  machine  delivering 
power  to  a  balanced  load.  If  the  E.M.F.  of  one  armature  be 
ea=Esmpt  and  the  current  be  ca=  C.sin  (pt—  \)  the  corre- 
sponding quantities  for  the  other  will  be 

eb  =  E  cospt  and  cb  =  C  .  cos  (pt  —  X). 
Hence  the  instantaneous  power  will  be 

w  =  EC  {sin  pt  .  sin  (pt  —  X)  +  cospt  .  cos  (pt  —  X)}, 

which  reduces  to  w  =  S.fflcos  X,  or  a  constant  quantity.  Hence  in 
the  two-phase  machine  with  balanced  load  the  flow  of  power  from 
it  is  constant  ;  it  can  readily  be  shown  that  the  same  is  true  for 
the  three-phase  machine. 


CHAPTER  XIII. 

POLYPHASE  TRANSFORMATIONS. 
TRANSFORMATION    WITH   UNALTERED    PHASES. 

As  in  the  case  of  monophase  currents,  transformers  can  be  used 
for  the  purpose  of  changing  the  pressure  from  that  of  supply  to 
any  other  desired  value.  Owing  to  the  greater  number  of  circuits 
there  is  much  variety  possible  in  the  different  connections.  In  the 
case  of  the  two-phase  circuit  all  that  is  necessary  is  to  provide 
each  phase  with  a  transformer  of  the  required  ratio.  With  three- 
phase  currents  many  methods  of  connection  are  possible.  In  what 
follows  we  will  use  the  letter  $P  to  denote  the  virtual  pressure 
between  any  pair  of  mains  connected  to  the  primary,  and  y  for 
the  pressure  between  two  adjacent  mains  in  the  secondary.  The 
symbol  p  will  be  used  for  the  ratio  of  transformation  in  the 
different  transformers  used ;  it  will  mean  the  ratio  between  the 
secondary  pressure  in  a  coil  homologous  to  a  similarly  situated 
primary  coil,  and  the  pressure  in  that  primary  coil. 

When  the  transformation  is  star  to  star,  we  have  the  arrange- 
ment shown  in  Fig.  130,  where  for  convenience  the  primary  and 


Fig.  130. 


POLYPHASE  TRANSFORMATIONS 


157 


secondary  coils  are  shown  separately,  though  of  course  corre- 
sponding coils  are  in  fact  wound  on  the  same  iron  core.  It  is 
evident  that  in  this  case  the  ratio  #7<f^  is  merely  p,  since  the 
connections  of  primary  and  secondary  are  the  same  in  form.  The 
vector  diagrams  are  shown  below,  the  lengths  of  the  lines  in  the 
same  can  be  taken  as  representing  the  corresponding  virtual 
values  of  the  pressures. 

If  the  transformation  be  mesh  to  mesh,  as  in  Fig.  131,  similar 
considerations  evidently  apply,  and  the  ratio  ^/^  is  again  p. 
In  this  case  it  will  be  seen  that  if  one  of  the  transformers  be 


Fig.  131. 

suppressed,  as  shown  to  the  right-hand  side,  no  difference  is 
produced  in  the  pressure  triangles.  Of  course  the  currents 
flowing  to  the  junctions  are  somewhat  altered  since  the  magnetising 
currents  of  the  transformers  now  come  down  only  two  mains,  and 
hence  balance  will  be  slightly  disturbed,  but  the  supply  of  energy 


Fig.  132. 


158 


ALTERNATING   CURRENTS 


from  the  secondaries  will  not  be  interfered  with.  This  is  some- 
times of  importance,  as  such  a  mesh  arrangement  will  permit  the 
temporary  cutting  out  of  one  of  the  transformers  should  circum- 
stances render  it  necessary. 

But  it  is  not  essential  that  both  primary  and  secondary  circuits 
should  have  the  same  connections,  the  former  may  be  connected  as 
a  star,  and  the  latter  as  a  mesh  as  in  Fig.  132.  An  inspection  of 
the  volt  diagram  below  will  show  that  the  ratio  ^/J5  is  now 
p/V3,  for  the  ratio  of  &  to  ^  is  now  p  while  that  of  &  to  ^ 
is  V3. 

Similarly  if  the  primaries  be  in  mesh  and  the  secondaries 
in  star  the  state  of  affairs  is  shown  in  Fig.  133.  Here  it  is  readily 
seen  that  #7^  is  p .  V3. 


Fig.  133. 

Two-  and  three-phase  transformation.  But  not  only  can 
we  transform  from  one  form  of  three-phase  connection  to  another, 
but  also  from  three-phase  to  two-phase.  Consider  two  primaries 
connected  as  shown  in  Fig.  134,  and  let  the  turns  in  the  primary  db 
be  V3/2  times  those  in  ac  while  the  point  d  is  the  centre  of  the 
winding  ac.  Let  two  equal  secondaries  be  provided  to  these 
primaries  as  shown,  and  let  three-phase  currents  be  supplied  to  the 
primaries,  we  shall  see  that  the  secondaries  will  have  two-phase 
relation  between  the  pressures.  Let  the  equilateral  triangle  ABC 
be  drawn  and  let  BD  be  the  perpendicular  from  B  on  AC.  Then 
BD  is  V3/2  of  any  of  the  sides,  and  the  three  lines  forming  the 
sides  are  at  120° ;  hence  this  figure  will  represent  the  state  of 
pressures  in  the  primaries  of  the  two  transformers.  It  follows 
that  the  pressures  induced  in  the  two  secondaries  will  have  a  phase 
relation  corresponding  to  that  of  the  lines  AC  and  BD,  that  is, 
they  are  in  quadrature.  If  the  two  secondaries  have  such  a 


POLYPHASE  TRANSFORMATIONS 


159 


number  of  turns  in  them  as  to  produce  equal  pressures,  they  will 
form  an  ordinary  two-phase  system.  The  ratio  of  #7<^  will 
evidently  be  that  corresponding  to  the  primary  ac  and  its  related 
secondary.  The  other  secondary  will  have  the  same  number 
of  turns,  but  the  primary  will  need  the  proper  number  of  turns 
mentioned  above. 


-Xooiro  o  o  a  op 


y> 


It  is  evident  that  if  two-phase  current  be  fed  into  such  a  pair 
of  transformers  in  the  reverse  way,  three-phase  currents  will  be 
delivered  from  the  other  terminals. 

The  above  method  necessitates  a  special  pair  of  transformers ; 
should  it  be  necessary  temporarily  to  use  transformers  with  more 
ordinary  ratios  of  transformation,  the  following  is  an  approximate 
way  of  obtaining  two-phase  current  from  a  three-phase  supply. 


160 


ALTERNATING   CURRENTS 


Let  the  primaries  of  the  transformers  be  connected  in  mesh  on  the 
mains,  and  let  two  of  the  secondaries  be  joined  in  the  ordinary 
manner,  but  divide  the  winding  of  the  other  in  its  mid-point,  and 
connect  the  ends  in  the  opposite  way  to  that  ordinarily  used, 
as  shown  in  Fig.  135.  The  volt  diagram  for  the  primary  will  be  as 
shown,  that  for  the  secondary  will  have  two  of  the  vectors  for  the 
ordinarily  connected  secondaries  drawn  in  as  usual,  but  the  vectors 
for  the  others  will  consist  of  two  equal  halves  drawn  opposite  to 
the  normal  direction.  It  follows  that  the  volt  diagram  for  the 
secondaries  will  be  as  on  the  right.  Consider  that  figure,  and 
note  that  in  the  triangle  CAE  the  side  CE  is  equal  to  the  line 
AD  while  the  side  AE  is  \/3/2  of  that  line,  hence  the  tangent  of  the 
angle  ACE  is  \/3/2,  from  which  it  follows  that  that  angle  is  about 
40|°,  and  hence  the  angle  CAB  is  about  99°.  Thus  the  pressures 
represented  by  AB  and  AC  are  approximately  in  quadrature.  The 
ratio  of  &I&*  is  found  as  follows;  the  ratio  of  AD  to  «^is  p,  while 
it  will  be  seen  that  ^is  V7/2  times  AD,  hence  <97<^is  V7/2 .  p. 

Transformation  to  six  or  more  phases.  We  will  now 
consider  how  it  is  possible  to  obtain  from  a  three-  or  two-phase 
system  another  system  which  has  six  or  more  phases.  Such  a  case 
will  arise  when  we  consider  the  connection  of  rotory  converters 
to  a  polyphase  system,  but  for  the  present  purpose  the  system  of 
loads  to  which  such  a  transformer  is  attached  may  be  taken  as 
consisting  simply  of  ordinary  resistances  connected  on  to  the 


Fig.  136. 

several  secondary  terminals.  For  the  present  purpose  such 
circuits  should  be  taken  as  being  similar  to  one  another.  Let  P, 
Fig.  136,  be  a  star  connected  primary,  and  let  two  similar  secondaries 
S,  Slt  be  provided  to  each  transformer;  let  the  first  set  of  these  be 
connected  in  a  star,  as  is  also  the  second  set,  but  let  the  opposite 
set  of  ends  be  connected  to  the  star  centre  in  the  latter  case,  and 


POLYPHASE  TRANSFORMATIONS  161 

then  let  the  two  stars'  centres  be  joined,  as  shown  in  the  figure. 
The  representation  of  the  pressures  existent  in  the  secondaries 
thus  connected  is  evidently  given  by  the  hexagon  shown.  Thus 
from  the  given  three-phase  primary  we  have  derived  a  six-phase 
secondary.  The  ratio  &Jf  will  be  given  by  />/\/3,  for  &  is 
evidently  equal  to  the  pressure  at  the  terminals  of  any  of  the 
secondary  circuits,  and  hence  ^/^  is  p,  also  «^is  V3^i,  hence  the 
result  follows. 

The  primary  could  have  equally  well  been  connected  in  the 
mesh  fashion  preserving  the  double  star  for  the  secondaries,  this 
would  only  produce  a  different  value  for  9*j£P. 

The  same  result  can  be  obtained  with  a  double  mesh  for  the 
secondary  as  shown  in  Fig.  137.  The  primary  being  assumed 
meshed,  it  is  provided  with  two  secondary  windings,  each  being 
also  meshed,  but  the  ends  of  the  second  winding  are  reversed  as 


Fig.  137. 

shown.  If  the  load  to  which  such  a  set  of  secondaries  is  joined  be 
such  as  to  require  symmetry  in  each  of  six  circuits,  the  vector 
representation  is  as  given.  It  will  be  noted  that  while  the  star 
case  of  necessity  produces  a  definite  neutral  point,  in  this  case  the 
neutral  is  fictitious  and  depends  on  the  form  of  the  load.  The 
ratio  &\gP  is  p/\/3,  for  we  have  ^  is  \/3  times  9*  and  also  #J  is 
p  times  «^.  As  in  the  last  case  a  star  connected  primary  could 
have  been  employed,  with  alteration  of  the  ratio  &/&. 

As  an  example  of  the  flexibility  of  a  polyphase  system  of 
transformation  we  will  now  see  how  a  twelve-phase  winding  can  be 
derived  from  a  two-phase  one.  Let  the  primaiy  be  connected  to 
a  two-phase  set  of  mains,  as  in  Fig.  138,  and  let  each  transformer 
be  provided  with  five  secondaries  forming  sets  Si  and  $2;  let  the 
turns  in  the  secondary  Dd  of  Si  be  #  and  wind  the  others  as 
follows :  Cc  and  Ff  with  V3/2 .  x  turns  and  Bb  and  Gg  with  x/2 
turns.  Let  the  secondaries  of  S2  be  similarly  wound  as  shown  at 
Aa,  etc.  Consider  the  figure  below  the  secondaries  and  let  them 
be  connected  together  as  there  indicated  by  the  corresponding 

L.  11 


162 


ALTERNATING   CURRENTS 


letters,  the  whole  set  being  connected  at  the  twelve  points  on  the 
circle  to  an  appropriate  load  as  in  the  double  mesh  case  just 
considered.  The  length  of  li  or  Dd  will  be  proportional  to  x ;  it 
will  be  noted  that  each  of  the  sides  of  the  figure  subtends  an 
angle  of  30°  at  the  centre,  from  which  it  will  be  readily  seen,  by 
following  the  dotted  lines,  that  the  lines  joining  GE  to  Be,  Og  to 
ob,  Ca  to  uc,  and  FA  to  Uf  are  each  V3/2  times  H ;  also  the  lines 
joining  GE  to  Og,  Be  to  ob,  FA  to  Ca,  and  Uf  to  uc  are  each  one 
half  of  li.  Thus  the  lines  fulfil  the  relations  of  magnitude  and 
phase  relation  demanded  by  the  windings  of  the  secondaries,  and 


82 


s, 


C  » 


.x-OUUJUUUULJL|  lAAJUUUUUiy 


f 

D    •TrrrTTcrrrvYvvvTrr*  d 

C  •nrrrrTTTTTTnrnrir«  C 

B  Tvr»TnnrffTnnr-  b 


1   l  u 
e    .  o 

i 


"1 

n 

G 

/ 

t 

^53^ 

& 

dif 

V 

\ 
\ 

^***^ 

\ 
\ 
^v  \ 

Xv\ 

£ 

\ 

CcT 

xx 

s? 

s'    1 

I 
1 

/ 

uc 

Bt 

1 

1 

Yt> 

h^-J 

^-^^ 

Fig.  138. 

the  figure  represents  the  distribution  of  pressures  due  to  the  same. 
If  ^denote  the  pressure  between  adjacent  sides,  it  will  be  seen  that 
^is  li  multiplied  by  cos  75°,  and  if  p  denote  the  ratio  between 
the  principal  secondaries  li  or  Dd,  and  the  pressure  across  the 
outside  mains  of  the  primary  or  ^  we  have  £f/&  is  p  cos  75°. 


Case  of  a  common  iron  core.  In  the  connections  de- 
scribed each  transformer  was  taken  to  be  a  separate  piece  of 
apparatus,  but  in  some  cases  this  need  not  necessarily  be  the  case. 
The  magnetic  circuits  of  the  three  transformers  can  be  joined  by 
common  limbs  and  since  the  fluxes  will  be  out  of  phase  with  one 


POLYPHASE.  TRANSFORMATIONS  163 

another  in  those  limbs,  the  section  of  iron  employed  in  such  parts 
of  a  transformer  where  the  fluxes  add  can  be  of  less  cross  section 
than  when  separate  ones  are  used;  this  results  in  a  saving  of 
weight  and  space  and  some  diminution  of  core  loss.  In  cases 
where  a  breakdown  of  the  apparatus  connection  of  the  trans- 
formers necessarily  results  in  their  being  put  out  of  operation, 
such  a  construction  is  desirable,  but  in  such  cases  as  the  double 
mesh,  where  two  transformers  can  carry  the  load  when  one  of  the 
three  is  out  of  action,  the  common  magnetic  circuit  cannot  be 
used. 


11—2 


CHAPTER  XIV. 


THE  EOTATING  FIELD  OR  INDUCTION  MOTOR, 


The  rotating  field.  We  will  now  consider  the  most  important 
property  possessed  by  certain  polyphase  circuits,  which,  indeed,  is 
one  of  the  chief  factors  that  has  determined  their  use.  Consider 
the  case  shown  in  Fig.  139  where  two  similar  symmetrical  coils  A 
and  B  are  placed  with  their  axes  at  right  angles  and  their  centres 
coincident,  and  let  two  equal  alternating  currents  differing  by  90° 

B 


B 

Fig.  139. 

in  phase  be  flowing  in  those  coils.  A  will  produce  an  alternating 
magnetic  flux  at  the  centre  of  the  coil,  in  the  direction  of  the  arrow 
a,  and  if  the  current  flowing  be  simple  harmonic  so  will  be  this  field. 
In  the  same  way  the  coil  B  will  produce  a  simple  harmonic  field  in 
the  direction  of  the  arrow  6.  But  these  two  equal  fields  differ  in 
angular  position  by  90° ;  hence  the  resultant  field  produced  must 
be  representable  by  the  addition  of  two  equal  simple  harmonic 
motions  at  right  angles  in  space  and  in  quadrature  in  time.  But 
we  know  that  the  result  of  such  a  combination  is  a  uniformly 
rotating  quantity  which  has  a  constant  magnitude  of  the  same 


THE   INDUCTION   MOTOR 


165 


value  as  the  maximum  of  the  components,  and  rotates  once  for 
each  alternation.  This  follows  from  the  fact  that  a  uniformly 
rotating  motion  is  equivalent  to  two  equal  simple  harmonic  ones 
along  axes  at  right  angles  differing  in  phase  by  90°. 

For  let  OP,  Fig.  140,  be  a  line  of  constant  length,  A,  rotating  with 
constant  angular  velocity,  p,  about  0 ;  let  Ox  and  Oy  be  the  usual 
perpendicular  axes,  and  let  the  position  of  the  point,  P,  be  specified 


Fig.  140. 

by  the  angle  0  reckoned  from  Ox.  Projecting  OP  on  the  axes  it 
is  seen  that  the  vector  OP  is  equivalent  to  the  two  vectors  OA 
and  OB.  But  we  have  OA  =  OP. cos  0,  and  OB  =  OP.  sin  0: 
further  6—pt,  thus 

OA  =  A  sinpt,     OB  =  A  .  cospt  =  A  .  sin  (pt  —  5- ) . 

V         A  / 

Thus  the  uniform  circular  motion  of  OP  is  equivalent  to  two 
simple  harmonic  motions  along  perpendicular  axes,  of  the  same 
amplitude  as  the  length  of  the  rotating  quantity  but  differing  in 
phase  by  90°,  that  is,  in  quadrature  both  in  space  and  time; 
hence  the  converse  holds  true. 

The  magnetic  field  at  the  centre  of  the  coils  will  thus  be  a 
rotating  one  of  constant  strength.  If  the  two  fields  are  not  at 
right  angles  both  in  time  and  space,  or  if  they  are  unequal  in 
magnitude,  the  field  produced  will  in  general  be  still  a  rotating  one, 
but  instead  of  being  of  constant  strength  it  will  vary  in  strength 
during  the  revolution  and  will  in  fact  be  an  elliptical  harmonic 
field ;  when  the  time  phase-angle  is  zero,  that  is,  the  two  currents 
are  cophased  in  time,  the  field  is  of  course  stationary  but  alternating. 

A  similar  result  can  be  shown  to  occur  if  instead  of  two  equal 
fields  at  right  angles  in  time  and  space  we  have  three  such  equal 
fields  at  angles  of  120°  in  time  and  space.  Let  the  coils  be 
disposed  as  shown  in  Fig.  141  and  let  three  currents  be  flowing  of 


166 


ALTERNATING   CURRENTS 


magnitudes  such  as  to  produce  a  field  of  maximum  strength  B  at 
the  centre  of  either  of  the  three  coils.     Considering  the  field  of  A 


B 


as  the  standard  of  reference  and  writing  it  a=  B  sinpt,  the  fields 
of  B  and  C  will  be 

Cf=Bsin(^  +  120°),   and   c  =  B  sin  (  pt  +  240°). 

Substituting  for  the  sines  and  cosines  of  the  constant  angles  these 
become 

C/q  \ 

J  smpt  —  —  cos  pt\  , 


and 


f  /^  \ 

c  =  -  B  U  sin^  +  -^  cos  ptj  . 


Now  take  as  standard  directions  for  resolution  that  of  the  field  a 
and  the  perpendicular,  and  let  x  and  y  be  the  resolved  components 
of  a,  b  and  c  on  these  lines  :  we  have 

#=a  +  c.cosl20°  +  c.cos2400,   and   y  =  b.  sin  120°  +  c.  sin  240°. 
On  substitution  and  reduction  these  expressions  become  : 
x  =  f  B  sin  pt   and   y  =  —  f  B  cospt. 

Hence  if  Br  be  the  resultant  of  the  three  vectors,  and  6  be  the 
angle  it  makes  with  oy,  we  have 


:  tan#  =  -, 

y 


which  gives 


=t,   or 


Thus  the  resultant  field  is  one  of  constant  strength,  l£  times  each 
alternating  field,  rotating  with  an  angular  velocity  equal  to  27rw, 
where  n  is  the  periodicity  of  the  currents.  It  should  be  noted 


THE   INDUCTION   MOTOR  167 

that  the  sign  of  the  angular  velocity  is  dependent  on  the  order  in 
which  the  different  alternating  fields  grow  in  strength. 

Resulting  torque.  Let  a  small  metal  disc  be  pivoted-  on  an 
axis  which  is  perpendicular  to  this  rotating  flux,  at  the  centre ^of  the 
coils.  The  flux  will  cut  this  disc  and  hence  currents  will  be  produced 
in  it ;  between  these  currents  and  the  flux  there  will  be  a  reaction 
resulting  in  a  couple  being  produced  tending  to  rotate  the  disc. 
If  there  be  any  opposing  couple  acting  on  the  disc,  for  example 
one  due  to  friction,  it  will  evidently  run  at  such  a  speed  that  the 
couple  produced  by  cutting  the  rotating  field  will  be  exactly  equal 
to  this  opposing  couple ;  if  the  opposing  couple,  to  whatever  it  be 
due,  is  zero,  the  disc  will  run  as  fast  as  the  field  rotates. 

The  rotating  field  or  induction  motor.  The  above  arrange- 
ment constitutes  a  simple  form  of  motor  but  it  would  produce  only 
a  small  couple  since  the  flux  in  the  air  is  necessarily  small.  If  by 
any  means  we  can  arrange  that  the  different  fluxes  are  produced 
in  an  iron  circuit,  much  larger  couples  would  be  attainable ;  the 
following  arrangement  enables  this  to  be  done. 

Let  two  sets  of  stampings  be  provided  of  the  form  shown  in 
Fig.  142,  both  inner  and  outer  having  holes  or  slots  to  carry  wind- 
ings, the  inner  set  being  all  rigidly  fixed  to  an  axis,  and  capable  of 
rotation.  Let  the  outer  set  be  divided  into  four  equal  parts  as 
shown  by  the  dotted  lines  and  wind  two  opposite  parts  as  shown 


Fig.  142. 

in  the  figure,  where  for  simplicity  only  one  set  of  windings  is 
indicated.  If  an  alternating  current  be  passed  through  this 
winding  it  will  produce  an  alternating  flux  as  shown  by  the  lines 
crossing  the  small  air  gap  that  is  left  between  the  two  sets  of 
stampings.  In  the  same  way  a  second  set  of  windings  placed  in 
the  holes  that  are  in  the  other  part  of  the  circumference  will 


168 


ALTERNATING   CURRENTS 


produce  a  flux  distribution  at  right  angles  to  the  above.  Let  the 
two  currents  supplied  be  in  quadrature  in  time  as  shown  by  the 
curves  A  and  B  in  Fig.  143,  and  consider  the  eight  successive  points 
in  one  period  indicated  thereon.  The  distributions  of  current  in 
the  two  sets  of  windings  for  these  different  points  are  roughly 
shown  in  the  eight  circles  below,  the  dots  and  crosses  indicating 
in  the  usual  manner  the  direction  of  the  currents,  and  the  number 
of  them  affording  an  indication  of  the  current  strength  in 


Fig.  143. 

the  coils.  The  direction  of  the  flux  produced  by  the  belts  of 
current  will  be  as  indicated  in  each  case  by  the  central  arrow,  and 
it  will  be  seen  that  this  arrow  executes  one  rotation  for  one 
complete  alternation  of  the  currents.  Thus  such  an  arrangement 
would  produce  a  rotating  belt  of  flux  the  angular  velocity  of  which 
is  such  that  the  number  of  revolutions  per  second  made  by  it  is 
equal  to  the  periodicity  of  the  currents  flowing  in  the  windings. 

Slip  and  torque.  Up  to  the  present  nothing  has  been  said 
as  to  the  use  of  the  holes  left  in  the  interior  set  of  discs.  Let  each 
of  these  have  threaded  through  it  a  bar  of  copper,  and  let  the  two 
sets  of  ends  of  these  bars  be  carefully  soldered  to  two  complete 
rings,  then  sets  of  closed  electric  circuits  are  formed  in  which 
pressures,  and  consequently  currents,  can  be  produced  by  any 
changing  flux.  Thus  the  rotating  field  which  will  rush  round  past 
these  rods  will  induce  currents  in  them  and  hence  a  torque  will  be 
produced  by  the  reaction  between  the  induced  currents  and  the 


THE   INDUCTION    MOTOR  169 

rotating  field,  tending  to  turn  the  interior  set  of  stampings  round. 
The  outer  set  of  stampings  is  usually  at  rest,  and  hence  the  two 
sets  are  respectively  called  the  stator  and  rotor,  that  is,  the 
alternating  currents  are  fed  into  the  different  phases  of  the  stator 
and  the  rotating  field  thereby  produced  tends  to  turn  the  rotor 
round.  If  there  was  no  opposition  to  the  rotation  of  the  latter  it 
is  evident  that,  as  in  the  case  of  the  disc,  it  would  run  up  to  such 
a  speed  that  no  couple  was  produced,  in  other  words  to  such  a 
speed  that  the  rods  in  the  rotor  had  no  currents  in  them  and 
did  not  cut  the  flux  produced  by  the  stator.  This  means 
that  it  would  run  at  the  same  angular  velocity  as  the  stator's 
field  or  in  synchronism  therewith.  If,  on  the  other  hand,  there 
be  any  couple  acting  on  the  rotor  tending  to  oppose  its  motion, 
the  rotor  would  have  to  produce  a  couple  equal  to  this  and  would 
therefore  have  to  cut  the  field  of  the  stator,  which  means  that  it 
would  run  more  slowly  than  in  synchronism ;  the  greater  the 
couple  required  the  greater  would  be  the  difference  between  the 
angular  velocity  of  the  rotating  field  and  that  of  the  rotor.  If  H 
be  the  angular  velocity  of  the  field  and  o>  that  of  the  rotor  the 
difference  between  them  will  be  <r  =  II  —  o>.  The  quantity  <r  is 
often  called  the  Slip  of  the  rotor,  and  this  slip  would  increase 
with  increased  demand  for  torque. 

Form  of  stator  winding.  In  Fig.  142  the  two  phases  there 
considered  are  shown  for  convenience  as  being  wound  on  distinct 
parts  of  the  stator ;  it  must  not  be  assumed  that  this  is  necessarily 
the  case  in  practice.  The  wires  forming  the  two  sets  of  windings 
can  be  distributed  in  any  desired  symmetrical  manner,  for  example, 
each  may  occupy  half  the  circumference  instead  of  one  quarter  as 
shown.  All  that  is  essential  is  that  the  fields  produced  by  the 
two  should  be  the  same  for  the  same  current,  and  that  the  space 
relation  should  be  one  of  quadrature.  The  effect  of  different  forms 
of  such  winding  is  solely  to  produce  different  shaped  distributions 
of  flux  in  the  air  gap.  For  convenience  in  description  and  clearness 
in  the  figures  the  simplest  one  is  taken,  but  it  must  not  be 
assumed  it  is  an  ordinary  form. 

The  rotating  belt  of  flux.  We  see,  then,  that  when  a 
couple  is  demanded  from  the  rotor,  in  addition  to  the  currents 
flowing  in  the  stator  there  will  be  currents  in  the  conductors  of 
the  rotor,  and  that  the  latter  will  rotate  at  a  somewhat  lower 
velocity  than  the  field  does,  the  difference  being  just  enough  to 
cause  the  currents  induced  in  the  rotor  to  be  of  the  proper  amount 
to  produce  the  required  couple.  The  currents  in  the  individual 
windings  of  the  stator  and  rotor  must  also  in  this  case  evidently 
combine  in  effect  so  as  to  be  equivalent  to  bands  of  currents  which 
may  be  looked  on  as  rotating  in  the  same  way  as  the  field  does, 
and  in  order  to  make  the  case  possible  of  treatment  by  a  graphical 


170 


ALTERNATING    CURRENTS 


method  it  is  necessary  to  idealize  it  in  some  manner  so  that  the 
different  quantities  can  be  represented  by  vectors. 

The  band  of  flux  that  is  rotating  will  at  any  moment  have 
some  specified  distribution  in  the  air  gap  and  the  simplest  case  to 
take  will  be  to  assume  that  it  is  so  distributed  that  we  can  repre- 
sent its  intensity  by  the  ordinates  of  a  sine  curve.  Thus  if  the 
horizontal  line  (Fig.  144)  be  the  circumference  of  the  air  gap  we 
may  consider  the  flux  at  any  moment  to  be  given  by  the  ordinates 


Direction  of  Motion  of 
the  Be  It  of  Flux. 


Fig.  144. 

of  the  sine  curve  drawn  there  and  if  this  sine  curve  be  imagined 
to  be  moving  bodily  with  the  velocity  that  the  rotating  field 
possesses  we  shall  have  a  simple  representation  of  the  state 
of  the  rotating  band.  We  must  now  see  under  what  conditions 
this  state  of  things  can  be  legitimately  represented  by  a  vector. 
In  Fig.  145  is  indicated  the  air  gap  and  its  band  of  flux, 


Fig.  145. 

the  thickness  of  the  shaded  area  outside  and  inside  the  air  gap 
circle  being  intended  to  indicate  the  density  of  the  field,  and 
being  such  as  to  correspond  with  the  ordinates  of  Fig.  144. 
Let  the  line  OF  be  drawn  from  the  centre  to  the  point 
where  the  maximum  value  of  the  field  occurs,  then  if  this  line 


THE    INDUCTION   MOTOR 


171 


rotate  with  the  field  it  can  be  taken  to  represent  the  whole 
distribution  of  flux  in  the  gap,  its  length  being  taken  so  as  to 
measure  on  some  desired  scale  the  maximum  value,  4>,  of  the  flux 
passing  across  a  strip  in  the  air  gap  one  centimetre  in  breadth. 
The  difference  between  this  form  of  vector  representation  and  our 
old  one  must  be  noted.  Here  there  is  no  line  considered  on  which 
this  vector  is  projected ;  it  must  be  looked  on  as  a  line  of  constant 
length  rotating  with  the  angular  velocity  of  the  field,  its  position 
pointing  to  the  place  where  the  flux  is  a  maximum.  It  is  possible, 
then,  to  represent  directed  quantities  which  are  distributed  in 
space  according  to  a  sinusoidal  law  provided  their  representative 
vectors  are  drawn  in  accordance  with  the  above  conventions. 

Composition  of  alternating  fluxes.  This  rotating  flux  is 
due  to  the  fact  that  two  stationary  alternating  fluxes  are  co-existing 
at  the  same  time  in  the  stator  with  a  phase  difference  of  90°. 
Consider  the  case  of  a  two-phase  stator,  and  assume  that  we 
replace  the  actual  flux  distribution,  which  at  any  instant  exists  in 
the  air  gap  due  to  the  action  of  one  set  of  windings,  by  a  sine 
distribution  as  shown.  If  at  the  instant  considered  the  latter 
flux  per  centimetre  breadth  have  a  maximum  value  <f>  (Fig.  146), 
the  expression  for  the  flux  at  any  point  P  given  by  the  angle  6 


Equivalent  S/ne 
flux.pha.se  A . 


0 j 


Circumference  -2n 


Equi  v a  lent  Sine 
f I  UK,  phase  B. 


Fig.  146. 

reckoned  from  a  fixed  point  in  the  stator  and  due  solely  to  the 
phase,  A,  will  evidently  be  given  by  <f> .  sin  6.  But  this  flux 
is  varying  with  the  current  flowing  in  the  windings  of  A 
and  we  can  assume  that  it  varies  with  time  in  a  sinusoidal 
manner  also  and  thus  the  maximum  is  given  by  <f>  =  4>  sin  pt . 
Hence  the  flux  at  any  point  P  due  to  the  phase  A  will  be 


172  ALTERNATING   CURRENTS 

given  by  <1>  sin  pt .  sin  6  at  the  angle  6  and  the  time  t.  The  other 
phase  B  is  so  wound  that  its  flux  attains  its  maximum  at  a  point 
one  quarter  further  round  the  circumference  and  hence  will  be  given 
in  space  by  the  cosine  curve  shown  below.  The  maximum  attain- 
able value  of  the  flux  will  be  the  same  as  for  phase  A  but  will 
not  be  attained  till  a  quarter  period  after  A,  hence  the  field  con- 
tributed by  B  at  the  same  point  P  will  be  given  by  <!>  cos  pt .  cos  0. 
The  flux  in  the  gap  will  then  be  given  by  the  sum  of  these 
or  by  4>  (sin  pt .  sin  9  +  cos  pt .  cos  0).  By  ordinary  reduction  this 
becomes  <E>cos(#—  pt).  It  follows  that  the  position  of  the 
maximum  flux  is  given  by  cos(#—  pt)  =  l  or  6—pt=Q,  that  is 
6=pt,  or  that  the  position  of  this  maximum  is  defined  by  the 

equation  —=p,  hence,  if  H  be  the  angular  velocity  of  the  field,  we 

have  O  =  p.  Thus  the  maximum  flux,  with  its  accompanying 
sinusoidal  band  of  flux,  will  fulfil  the  required  condition  of  rotating 
with  uniform  velocity  corresponding  to  the  periodicity,  unchanged 
in  shape,  provided  the  assumed  conditions  are  fulfilled,  namely,  the 
stationary  alternating  fluxes  that  produce  it  must  be  sinusoidal  in 
space  distribution  in  the  air  gap,  and  their  maxima  must  be^  equal 
simple  harmonics,  both  space  and  time  differing  by  J  period  in 
each. 

It  may  be  noted  in  passing  that  a  similar  proof  can  be  given 
for  the  three-phase  winding.  In  this  case  the  stationary  flux  due 
to  each  stator  winding  will  be  severally  given  by  the  three  expres- 
sions 

<E>  sin  6  sin  pt, 

<&  sin  (0  +  |TT)  (sin ^  +  ITT), 
and  <£  sin  (6  +  4  TT)  (siupt  +  f  TT). 

The  flux  in  the  gap  will  be  the  sum  of  these.  It  is  left  to  the 
student  to  expand  and  add  the  above.  It  will  be  found  on  so 
doing  that  each  term  contributes  two  terms  to  the  sum,  in  each 

case  one  of  these  terms  is  -^-cos(#  —  pt),  and  the  other  three  will 

cancel  on  addition,  hence  the  rotating  flux  will  have  a  maximum 
value  which  is  1^  times  either  of  the  components,  and  rotates  at  a 
number  of  revolutions  per  second  equal  to  the  periods  per  second 
made  by  the  impressed  currents. 

For  the  future  consideration  of  the  case  we  will  take  for  simpli- 
city the  case  of  a  two-phase  stator,  as  it  is  evident  that  the  state 
of  things  can  be  represented  by  the  same  constructions  in  the 
cases  where  the  rotating  flux,  etc.  is  the  same,  to  whatever  form  of 
winding  it  may  be  due. 

The  maximum  value  of  the  flux,  that  is,  <t>,  is  evidently 
approximately  equal  to  the  maximum  intensity  of  the  induction  in 


THE   INDUCTION   MOTOR 


173 


the  gap  multiplied  by  the  length  of  the  same  measured  parallel  to 
the  axis  of  the  stator. 

The  current  bands.  Ideal  winding.  The  next  point  is 
to  consider  the  nature  of  the  current  bands  that  must  exist  in 
order  that  this  sine  distribution  of  flux  may  be  possible. 

Since  the  rotating  field  has  been  assumed  to  have  a  sine 
distribution  in  space  it  follows  that  the  distribution  of  the  current 
band  or  bands  to  which  its  existence  is  due  must  follow  the  same 
law.  But  in  the  windings  of  both  rotor  and  stator  the  wires  carrying 
the  currents  forming  the  bands  are  arranged  in  such  a  manner 
that  the  current  at  any  instant  is  the  same  over  the  arc  belonging 


1 

A 

B 

A 

B 

B 


B 


Fig.  147. 

to  any  definite  coil.  Thus  at  the  instant  the  current  in  one  phase 
(B)  of  the  stator  is  zero,  and  hence  that  in  the  other  (A)  a  maxi- 
mum, the  current  distribution  round  the  gap  due  to  the  stator  will 
be  roughly  as  shown  at  I  in  Fig.  147,  at  a  quarter  period  after  it 
will  be  as  in  II,  and  at  another  quarter  period  as  in  III.  It  follows 
that  even  if  the  currents  supplied  follow  a  sine  variation  with  time, 
the  distributions  of  the  current  bands  in  the  circumference  of  the 
stator  cannot  do  so,  and  this  must  of  necessity  be  the  case  in  any 
practicable  form  of  winding.  A  similar  state  of  things  exists  in 


174  ALTERNATING   CURRENTS 

the  rotor,  so  that  in  neither  case  could  we  have  a  sine  distribution 
of  current  in  space  with  the  actual  windings  employed.  It  follows 
that  in  order  to  make  the  case  amenable  to  vector  treatment  we 
must  replace  the  actual  windings  by  ideal  windings  which  will 
enable  sine  distributions  of  current  to  exist,  but  will  at  the  same 
time  have  the  same  electrical  conditions  and  the  same  magnetic 
effects  as  those  actually  employed. 

This  ideal  winding  can  be  taken  to  have  the  following  form. 
Let  the  actual  winding  of  any  phase  of  the  stator  be  imagined 
to  be  replaced  by  one  made  up  of  a  very  great  number  of 
wires  very  close  together,  the  number  per  centimetre  of  the 
circumference  of  the  stator  being  so  arranged  that  they  vary  as 
the  ordinates  of  a  sine  curve  round  that  phase.  Then  any  current 
flowing  in  the  stator  will  very  nearly  produce  a  belt  of  current 
which  is  of  the  desired  sine  form  in  space.  In  order  that  the 
resistance  of  such  a  set  of  turns  may  correspond  to  that  of  the 
actual  winding  it  is  evident  that  it  must  also  be  arranged  that  the 
ideal  winding  has  the  same  resistance  for  each  centimetre  run  that 
the  actual  winding  has.  Thus  if  the  real  winding  has  a  resistance 
of  r  ohms  and  if  it  occupies  I  centimetres  of  the  circumference,  the 
equivalent  ideal  winding  should  have  a  resistance  per  centimetre 
run  of  the  constant  value  r/l  ohms.  This  would  evidently  entail 
that  the  gauge  of  its  wire  should  vary  continuously  from  one  point 
to  the  next,  but  there  is  no  difficulty  in  forming  a  conception 
of  this  state  of  affairs.  We  thus  have  a  winding  of  such  a  nature 
that  it  produces  a  sine  curve  of  current  distribution  round  the 
stator,  and  has  the  same  resistance  per  centimetre  as  the  true 
winding.  The  only  other  condition  that  must  be  fulfilled  is  that 
it  shall  produce  the  same  magnetic  effect,  and  this  can  be  secured 
by  imagining  that  the  total  ampere  turns  of  the  real  and  ideal 
winding  are  the  same.  Now  let  the  other  phase  have  a  similar 
winding  spaced  out  in  quadrature  with  the  first,  and  we  have 
a  form  of  winding  which  is  amenable  to  vector  treatment.  It  may 
be  objected  that  this  ideal  winding  is  so  far  removed  from  the 
actual  one  as  to  make  the  results  arrived  at  of  no  value,  but 
in  practice  it  is  found  that  the  results  deduced  from  the  con- 
sideration of  this  form  correspond  closely  with  those  obtained  by 
a  test.  The  rotor  must  in  the  same  way  be  imagined  to  be 
provided  with  a  winding  arranged  in  a  practically  continuous 
manner. 

With  such  windings  it  will  at  once  be  evident  that  all  that 
has  been  proved  about  the  combination  of  two  alternating  fields  at 
right  angles  in  time  and  space  will  at  once  follow  for  two  alternating 
currents  of  sine  form  with  time  when  supplied  to  these  windings. 
Hence  the  combination  of  the  two  alternating  currents  in  the 
stator  will  result  in  a  band  of  current  distributed  sinusoidally  in 
the  winding  and  rotating  in  the  stator  windings,  whose  maximum 


THE   INDUCTION   MOTOR 


175 


corresponds  with  that  of  the  maximum  of  the  two  alternating 
currents  in  the  two-phase  case.  The  necessary  phase  relation  that 
must  subsist  between  the  two  bands  of  current  in  the  rotor  and 
stator  and  the  band  of  rotating  flux  will  be  considered  on  p.  177. 

The  induced  E.M.F.s.  We  must  now  consider  the  E.M.F.S 
that  will  necessarily  be  produced  in  these  two  belts  of  wires  by 
the  rotating  flux,  and  also  the  methods  by  which  they  can  be 
represented.  Let  any  wire  in  the  stator  winding  be  considered 
(Fig.  148) ;  it  is  at  rest  and  the  flux  is  rushing  past  it  at  a  certain 
velocity,  ^ ,  which  will  be  very  nearly  equal  to  the  product  of  the 
angular  velocity  of  the  field,  fi,  and  the  radius  of  the  rotor,  say  p.  If 
<f)  is  the  value  of  the  flux  in  which  that  wire  stands  at  the  instant 


Vi  or(Vi-V2) 


A 


Fig.  148. 

taken,  there  will  be  an  E.M.F.  generated  in  it  of  the  amount  v^t 
hence  each  of  our  little  wires  will  have  similar  E.M.F.S  generated  in 
them,  and  thus  there  will  be  a  sine  distribution  of  induced  E.M.F.S 
in  the  stator  windings  due  to  the  rotating  field,  which  will  form 
a  band  of  induced  stator  E.M.F.  lying  exactly  in  phase  with  the  flux 
produced  in  the  air  gap.  The  maximum  value  of  this  E.M.F.  will 
evidently  be  4>.  vl}  and  the  vector  corresponding  to  it  will  be  of  such 
a  length  as  to  represent  the  E.M.F.  to  some  suitable  scale,  and  will 
point  in  the  same  direction  as  that  representing  the  band  of 
induction  in  the  gap.  Now  let  the  same  point  be  one  of  the  rotor 
wires ;  in  this  case  the  wire  itself  is  moving  in  the  same  direction 
as  the  flux  but  with  a  velocity  v2  which  is  equal  to  the  product 
of  the  radius  of  the  rotor,  p,  into  its  angular  velocity,  <w,  and  hence 
the  relative  velocity  of  the  wire  and  the  field  is  (vj  —  v2).  It 
follows  that  in  this  case  there  is  also  a  sine  distribution  of  E.M.F. 
in  the  rotor  wires,  but  its  maximum  will  be  only  ^>.(vl  —  v2).  The 
vector  representation  of  this  would  be  a  line  whose  length  inter- 
preted on  the  proper  scale  for  pressures  would  give  the  quantity 
<p  ,(vl  —  v2).  The  direction  of  that  vector  in  space  will  be  the  same 
as  the  last  one  and  the  same  as  the  gap-flux  vector.  This  some- 
times causes  the  student  a  little  difficulty  since  the  angular 
velocity  of  the  field  and  rotor  are  different,  but  it  must  be 
recollected  that  since  the  velocity  of  the  field  is  ^  and  that  of  the 
rotor  is  v2,  the  relative  velocity  of  the  two  must  be  the  difference. 


176  ALTERNATING   CURRENTS 

It  follows  that  while  the  band  of  E.M.F.  in  the  rotor  is  moving 
relative  to  the  wires  at  the  velocity  (^  —  v2)  it  must  be  moving 
in  space  at  the  velocity  (^  —  v%)  +  vz  or  vl}  and  hence  at  the  same 
velocity  as  the  other  E.M.F.  and  the  flux  vectors.  The  same 
consideration  evidently  applies  to  the  two  current  vectors.  Hence 
a  set  of  vectors  such  as  we  have  considered  will  form  a  definite 
geometrical  figure  which  can  be  considered  to  be  rotating  at  the 
velocity  v±  about  an  axis  and  will  then  represent  the  state  of 
things  in  the  motor. 

If  the  whole  machine  be  imagined  to  be  rotated  backwards 
with  the  velocity  of  the  field,  this  figure  will  be  reduced  to  rest  in 
space,  but  the  rotor  would  then  be  moving  backwards  with  the 
small  velocity  (vx  —  v2). 

The  impressed  pressure.  There  are  other  E.M.F.S  that  must 
be  taken  into  consideration,  and  in  particular  the  impressed  pressure 
at  the  terminals  of  each  phase  of  the  stator.  As  with  the  currents, 
these  will  necessarily  be  related  in  quadrature  both  in  time  and 
space,  and  with  our  idealized  band  of  stator  wires  must  be 
considered  as  constituting  a  rotating  belt  of  impressed  pressure 
in  the  constituent  wires  distributed  in  a  sinusoidal  manner  round 
the  circumference  at  any  moment  and  likewise  rotating  at  the 
velocity  v^.  This  band  will  evidently  have  a  definite  maximum 
value,  which  may  be  arrived  at  in  the  following  manner.  The 
virtual  value  of  the  pressure  applied  to  each  winding  is  known,  let 
it  be  S0y  then  the  maximum,  on  the  assumption  of  sine  variation 
with  time,  will  be  \/2  .  £0.  But  if  each  winding  on  the  stator  have 
t  turns  in  it,  the  maximum  E.M.F.  per  wire  will  be  \/2^0/£,  and  we 
may  take  this  as  the  value  of  the  maximum  of  the  sine  distribution 
of  pressure  that  is  existing  in  each  phase  of  our  ideal  stator.  We 
shall  shortly  have  to  deal  with  other  E.M.F.S  but  will  defer  con- 
sidering them  for  a  little. 

Phase  relations.  Up  to  the  present  nothing  has  been  said 
as  to  the  relation  between  the  two  current  bands  as  regards  phase. 
It  is  evident  that  if  the  flux  of  magnetism  did  not  require  any 
magnetising  force  to  produce  it  these  two  bands  would  at  every 
instant  exactly  exert  the  same  magnetic  effect  but  in  opposite 
directions,  and  would  consequently  be  represented  by  two  equal  and 
opposite  vectors.  But  to  force  the  flux  through  the  circuit  will 
require  a  certain  magnetomotive  force  which,  as  in  the  case  of  the 
transformer,  must  be  supplied  from  the  source  of  energy  giving 
current  to  the  stator.  When  the  details  of  the  magnetic  circuit 
are  given,  the  maximum  current  that  will  be  required  in  either 
phase  in  order  to  produce  any  desired  maximum  induction  in  the 
air  gap  can  be  calculated  in  a  similar  manner  to  the  transformer. 
Owing  to  the  presence  of  an  air  gap  this  maximum  current  is  far 
larger  than  in  a  transformer  of  about  the  same  size.  Hence  in 


THE   INDUCTION   MOTOR 


177 


our  ideal  winding  on  the  stator  another  sinusoidal  band  of 
current  having  this  ascertained  maximum  value  must  be  flowing, 
which  band  is  concerned  solely  in  producing  the  rotating  flux. 
Thus  the  actual  current  band  in  our  stator  will  consist  of  the 
combination  of  two  bands,  the  one  an  exactly  inverted  image 
of  the  rotor's  band  and  the  other  this  extra  magnetising  band. 
The  relative  phase  angles  of  their  representative  vectors  must 
now  be  found,  and  in  doing  so  it  must  be  borne  in  mind  that 
the  rotor's  band  of  current  is  due  to  an  E.M.F.  produced  by 
its  wires  cutting  the  rotating  flux.  Let  the  maximum  value 
of  the  flux  in  the  gap  be  as  before  <f>,  and  draw  the  vector  QF, 
Fig.  149,  from  the  point  Q  to  represent  the  magnitude  and 


position  of  the  maximum  of  the  flux.  Take  0  as  the  origin  for 
our  current  vectors,  then  in  any  one  of  our  rotor  wires  will  be 
induced,  as  we  have  seen,  an  E.M.F.  and  the  maximum  of  this  will 
be  4> .  (v-i  —  v2).  Thus  if  the  vector  OE2  be  drawn  of  this  length 
and  parallel  to  QF  it  will  represent  the  band  of  induced  E.M.F.  in 
the  rotor.  Owing  to  circumstances  which  will  be  shortly  gone 
into,  the  current  in  any  one  of  the  wires  of  the  rotor  will  lag  after 
this  E.M.F.,  by  a  small  angle  X.  This  angle  is  a  time  lag  angle 
for  the  rotor  current  after  its  pressure,  that  is  if  the  pressure  has 

its  maximum  at  a  definite  instant,  the  time  ^—  seconds  must 

ZTT 

elapse  before  the  current  is  a  maximum.  But  in  that  time  the 
flux  belt  will  have  advanced  through  the  same  angle  since  its 
velocity  is  such  that  it  makes  one  revolution,  2-Tr,  in  the  time  r, 
hence  to  represent  the  space  relation  of  the  rotor  current  we  can 
draw  its  vector  at  the  same  angle,  X,  to  the  air  gap  flux  vector, 

L.  12 


178 


ALTERNATING  CURRENTS 


that  is,  to  the  rotor's  E.M.F.  vector.  Hence  if  the  vector  OC2  be 
drawn  at  this  angle,  X,  to  OE<>  it  will  represent  the  current  band  in 
the  rotor.  From  the  method  in  which  the  winding  of  the  stator 
is  carried  out  it  will  be  seen  that  in  order  to  produce  a  band 
of  flux  which  can  be  represented  by  QF  we  must  have  a  belt 
of  current  acting  in  the  direction  shown  by  the  arrow  round  QF. 
But  such  a  band  is  to  be  represented  by  a  line  drawn  in  the 
direction  of  the  maximum  current  in  the  band,  and  hence  the 
vector  for  the  magnetising  band  must  be  drawn  from  0  perpen- 
dicular to  OEZ  or  QF,  as  shewn  by  00.  The  length  of  00  has  to  be 
taken  to  represent  on  the  scale  of  current  the  maximum  value  of 


Belt  of  Stator  current 


Belt  of  air  $ap  Flux 


Belt  of  Rotor  current 


Difference  of  current' 

that  is  belt  of 
magnetising  current 


Fig.  150. 


the  magnetising  current  band  referred  to  above.  Now  the  stator 
band  of  current  has  to  equilibrate  OC2  and  provide  00.  It  will 
therefore  be  given  by  the  parallelogram  drawn  in  the  figure  and 
will  be  represented  by  the  line  OClf  The  curves  in  Fig.  150  will 
give  an  idea  of  the  distribution  of  these  current  bands  round  the 
motor  at  any  moment :  in  addition  to  the  flux  band  and  the  two 
current  bands  a  lower  curve  is  drawn  which  is  the  difference 
between  the  two  latter,  and  hence  represents  the  band  of  mag- 
netising current,  it  will  be  seen  that  it  is  in  quadrature  with  the 
flux  band.  It  will  also  be  noticed  that  the  axes  of  the  maximum 
values  of  the  flux  and  the  two  main  current  bands  nearly 
correspond,  that  is  these  three  are  nearly  in  or  antiphase  with 
one  another. 

Leakage  fluxes.  Another  important  point  now  claims 
consideration.  Up  to  the  present  we  have  assumed  that  the  only 
flux  existing  is  that  which  passes  across  the  air  gap  and  is  cut  by 
the  conductors  both  of  the  rotor  and  the  stator.  But  the  currents 
in  the  two  sets  of  windings  can  produce  two  local  fluxes  round 
themselves,  which  fluxes  in  no  way  contribute  to  the  air  gap  flux. 
Consider  a  set  of  slots  which  is  situated  at  the  place  where  the  air 


THE    INDUCTION    MOTOR  179 

gap  flux  is  a  maximum,  this  flux  being  due  to  the  combined 
effects  of  the  wires  in  all  the  slots.  Then  from  what  we  have 
just  seen  the  two  currents  in  the  wires  in  the  slots  of  stator  and 
rotor  at  that  point  will  be  roughly  at  their  maximum  value  also. 
Hence  each  of  them  can  send  a  local  flux  round  the  wires  as 
shown  in  Fig.  151.  The  result  as  regards  the  magnitude  of  the  air 
gap  flux  is  manifestly  to  leave  it  unaltered  in  amount  but  it  will 
distort  its  distribution.  These  two  fluxes  are  of  exactly  the  same 


Sbztor 


^~^  ^ _.     — 


Rotor 


Fig.  151. 

nature  as  the  leakage  fluxes  in  the  transformer,  but  in  the  present 
case  they  will  be  much  bigger  in  proportion  than  in  any  trans- 
former, firstly  because  the  two  opposing  sets  of  windings  are  of 
necessity  placed  on  different  parts  of  the  magnetic  circuit,  and 
secondly  because  there  must  be  an  air  gap  between  them.  The 
leakages  will  be  less  in  proportion  the  smaller  the  air  gap  can  be 
made,  and  for  this  reason  the  gap  is  reduced  to  the  smallest  value 
consistent  with  safe  working.  Since  these  fluxes  have  by  far  the 
greater  part  of  their  path  passing  through  a  circuit  of  which  the 
reluctance  is  constant,  the  maximum  value  of  those  fluxes  will  be 
very  nearly  proportional  to  the  currents  that  severally  produce 
them,  so  that  we  can  write  them 

<|>i?1  =  ^1C1  and  3>«2  =  A:2C2,  ^  and  k2  being  constants. 

These  fluxes  will  show  their  presence  by  the  production  of 
additional  E.M.F.S  in  the  wires  both  of  stator  and  rotor  due  to  the 
fluxes  being  cut  by  the  wires  and  these  new  E.M.F.S  must  now  be 
considered  in  order  that  we  may  properly  complete  the  diagram  of 
our  ideal  motor. 

But  before  we  can  find  out  the  proper  directions  and  magnitudes 
of  the  vectors  for  those  E.M.F.S  it  is  necessary  to  consider  the  form 
that  the  leakage  flux  will  have  in  our  set  of  wires  carrying  the 
sinusoidal  band  of  current  in  either  set.  Let  the  curve  in  Fig.  152 
represent  either  band  of  current,  then  if  we  consider  two  wires  in 
the  winding  equally  distant  from  the  point  of  maximum  current, 

12—2 


180 


ALTERNATING   CURRENTS 


P,  the  fluxes  due  to  the  equal  currents  in  those  wires  will  flow  as 
shown  by  the  little  arrows  below  the  maximum,  and  since  these 
oppose  in  the  gap  there  will  be  no  nett  flux  embraced  by  the  wire 
at  P :  again  taking  two  wires  equidistant  from  the  point  of  zero 


Current 
Band 


Leakage 


Leakage. 
Flux. 


Fig.  152. 


current,  0,  the  directions  of  the  equal  local  fluxes  will  be  the  same 
again,  as  shown  by  the  little  arrows  below  the  minimum,  which  in 
this  case  run  in  the  same  direction ;  hence  the  maximum  leakage 
flux  will  occur  there,  that  is  at  the  point  where  the  current  is  zero. 
Hence  it  follows  that  each  of  the  bands  of  current  will  be  accom- 
panied by  a  band  of  leakage  flux  situated  in  space  at  right 
angles  to  the  position  of  the  band  and  having  a  maximum  value 
proportional  to  the  current  maximum  in  the  corresponding  current 
band. 

The  E.M.F.S  produced  by  the  wires  cutting  these  bands  will  be 
represented  in  the  usual  way  by  means  of  vectors  pointing  in  the 
same  direction  as  those  of  the  bands,  and  since  <tsl  and  <J>ffi  are 
the  values  of  the  maximum  flux  for  the  two  bands  of  leakage  flux, 
the  E.M.F.  produced  in  the  stator  wires  will  have  the  maximum 
value  vl .  Osl  while  that  in  the  rotor  wires  will  have  the  maximum 
value  (vi  —  v2)  <I>g2.  At  present  we  will  assume  that  the  values  of 
<J>gl  and  4>S2  are  known. 

Vector  Diagram.  We  can  now  proceed  with  the  full  dia- 
gram for  the  motor.  Take  Q,  Fig.  153,  as  the  origin  for  the  flux 
vectors  and  0  for  that  of  the  current  and  pressure  ones.  Draw  the 
line  QF  to  represent  <J>,  the  maximum  air  gap  flux.  Then  as  before 
if  OE2  be  O.fvj  — v2)it  will  represent  the  band  of  impressed  E.M.F. 
produced  by  the  gap  flux  in  the  rotor  wires,  and  if  OE^  be  equal  to 
4> .  V-L  it  will  represent  the  band  of  induced  E.M.F.  produced  in  the 
stator  wires  by  the  gap  flux.  If  r  denote  the  resistance  per  centi- 
metre run  of  the  rotor  band  of  conductors  and  if  C2  be  any  assumed 
value  of  the  maximum  current  in  them,  a  pressure  of  the  amount 
C2r  will  be  required  to  force  the  current  through  the  wire  as  far  as 


THE   INDUCTION   MOTOR 


181 


resistance  is  concerned.  Hence  this  pressure  will  be  required  in 
that  one  of  our  wires  which  is  at  the  point  of  maximum  current. 
In  addition  the  current  C2  will  have  produced  a  leakage  field  dis- 
tribution of  the  maximum  value  <&K,  just  proportional  to  its  own 
value,  but  the  vector  for  this  field  has  to  be  drawn  at  right  angles 


Fig.  153. 

to  that  for  the  current ;  also  this  flux,  <J>82,  will  induce  an  E.M.F. 
(^i  —  #2)  &K  in  our  rotor  wire,  the  vector  for  which  will  be  in  the 
same  direction  as  the  vector  for  O^.  It  follows  that  the  E.M.F.,  OE2, 
has  to  be  so  related  to  those  for  the  pressures  C2r  and  (^  —  v2) .  O^ 
that  it  is  the  vector  sum  of  the  two,  and  at  the  same  time  the  two 
component  vectors  are  at  right  angles.  Hence  if  we  draw  a 
semicircle  on  OEZ  and  make  Or  equal  to  C2r  the  other  side  rEz 
must  be  the  vector  representing  (vt  —  V2) .  4>^.  Let  this  be  drawn 
from  0  and  called  OS2,  then  the  length  OS2  will  be  the  value  of 
(#1  —  ^2)  ^«2  on  the  E.M.F.  scale.  It  follows  that  we  can  draw  the 
vector  QZ/2  parallel  to  OS2  to  represent  the  amount  of  the  leakage 
field  in  the  rotor.  Furthermore  if  the  triangle  QFR  be  drawn 
with  its  sides  parallel  to  those  of  the  triangle  E20r,  it  is  evident 
that  since  FR  will  be  parallel  and  equal  to  QL2,  the  line  QR  will 
represent  the  nett  flux  in  the  rotor  or  that  concerned  in  actually 
producing  the  pressure  that  impels  the  current  in  the  wires 
against  their  resistance  only.  We  now  proceed  as  on  p.  176 :  to 
produce  the  air  gap  flux  represented  by  QF  will  require  a  specia  1 


182  ALTERNATING   CURRENTS 

band  of  current  in  the  stator  found  as  described  on  p.  176,  and  its 
vector  will  be  0(7,  at  right  angles  to  QF.  The  Vector  representing 
the  rotor  band  of  flux  will  be  in  phase  with  Or,  let  it  be  given  by 
0(72.  Then  in  the  manner  before  described  we  can  find  the  vector 
representing  the  stator  current  band,  or  OC^  The  presence  of 
this  band  means  that  there  must  coexist  a  band  of  leakage  flux,  as 
we  have  seen,  and  the  vector  representing  this  flux  must  be  at 
right  angles  to  the  current  vector  and  of  a  length  proportional  to 
that  current.  It  is  drawn  at  QL^  Its  length  is  of  course  equal 
to  <3>S1  measured  on  the  flux  scale.  The  actual  total  flux  in  the 
stator  must  be  such  as  to  produce  both  QF  and  QLl}  and  will 
therefore  be  given  by  QS  where  SF  is  equal  and  parallel  to  QL^ 
The  existence  of  the  flux  QLt  in  the  stator  will  necessitate  the 
production  of  an  E.M.F.  of  the  amount  vl<&sl  in  a  direction  parallel 
to  QL1}  hence  the  line  0$j  can  be  drawn  to  represent  this  E.M.F. 

We  will  neglect  the  small  pressure  required  for  forcing  the 
currents  through  resistance  of  the  band  of  stator  wires  since  the 
values  of  the  induced  E.M.F.S  are  evidently  far  more  important. 
It  follows  that  the  impressed  pressure  band  referred  to  on  p.  176 
has  to  perform  two  functions,  firstly  to  equilibrate  OS1}  and 
secondly  to  equilibrate  the  E.M.F.  induced  in  the  wires  by  the 
air  gap  flux.  Hence,  if  we  draw  OE  equal  and  opposite  to  OE± 
and  OS  equal  and  opposite  to  08^  the  resultant  of  these  two 
vectors,  that  is  0V,  will  nearly  give  the  direction  and  magnitude 
of  the  band  of  impressed  pressure  in  the  stator. 

It  will  be  noted  that  the  flux  diagram  is  perpendicular  in 
space  to  the  current  one;  that  the  triangles  QFR  and  OE.2r  are 
similar,  the  ratio  of  their  sides  being  (^  —  v2),  and  the  triangles 
Q8F  and  OVE  are  similar,  the  ratio  of  the  sides  being  vlm 

Degree  of  approximation  in  sine  assumption.  A  con- 
sideration of  this  vector  figure  will  show  that  under  ordinary 
conditions  the  assumption  of  a  sine  band  of  flux  is  not  far  from 
the  truth.  For  it  will  be  seen  that  the  E.M.F.  due  to  the  rotating 
belt  of  flux,  that  is  the  E.M.F.  given  by  OE,  is  the  predominant  one 
in  the  stator.  If  the  impressed  pressure  curve  is  truly  sinusoidal 
in  each  phase,  it  must  follow  that  the  corresponding  E.M.F.  belt  is 
of  the  form  we  have  assumed,  namely  one  consisting  of  a  simple 
harmonic  curve  of  E.M.F.  rotating  in  the  stator  windings.  It 
necessarily  follows  that  the  E.M.F.,  OE,  will  also  be  practically  of 
the  same  form,  and  hence  that  the  flux  to  which  it  is  due,  that  is 
the  air  gap  flux,  is  also  a  belt  of  that  nature.  This  point  has  been 
experimentally  verified  by  direct  observation  on  the  instantaneous 
pressure  generated  in  test  coils  placed  in  the  air  gap,  and  it  was 
proved  that,  with  a  sine  wave  of  impressed  pressure,  the  rotating 
flux  is  sinusoidal  in  form  round  the  rotor.  The  total  stator  flux 
must  evidently  be  sinusoidal  under  such  conditions,  but  the  ex- 
perimental verification  would  be  more  difficult. 


THE    INDUCTION   MOTOR  183 

The  torque  of  an  induction  motor.  The  next  points  to 
be  considered  are  those  connected  with  the  operation  of  the 
motor  under  different  conditions  such  as  loaded,  starting,  etc. 
For  this  purpose  it  is  necessary  to  find  an  expression  for  the  torque 
that  such  a  motor  can  produce,  and  it  is  convenient  to  slightly 
alter  some  of  the  symbols  hitherto  used.  If  we  denote  as  before 
the  radius  of  the  rotor  by  p,  and  the  slip  (or  the  difference  between 
the  angular  velocity,  H,  of  the  field  and  that,  co,  of  the  rotor)  by 
o-,  we  can  write  vl  —  v2  =  p  .  <r.  Again  the  leakage  field  of  the  rotor 
is,  as  we  have  said,  proportional  to  the  current  in  our  wires  and 
if  &,  be  some  constant  depending  on  the  form  of  the  motor  the 
E.M.F.  induced  in  the  rotor's  wires  by  this  leakage  field  can  be 
written  k2  .  C2  .  par,  where  kz  .  C2  is  put  for  4>M.  But  for  any 
definite  rotor  the  quantity  k.2  .  p  is  fixed  and  we  will  denote  it  by 
the  single  letter  L.  Hence  if  the  triangle  OEp  (Fig.  153)  for  the 
pressures  existing  in  one  of  the  rotor's  wires  be  considered,  the 
sides  can  be  expressed  thus  :  OE.2  is  equal  to  oy><E>,  Or  to  C2r,  and 
Ezr  to  La-  .  C2.  It  follows  from  this  that  we  can  put 

CZV  or  C  = 


We  can  now  proceed  to  find  an  expression  for  the  torque 
exerted  between  the  rotor  and  the  stator.  Consider  the  instant 
when  the  air  gap  flux  has  its  zero  value  along  the  horizontal  line, 
OOj,  in  Fig.  154,  and  its  maximum  at  right  angles  along  QQl  :  at 
that  instant  the  current  band  in  the  rotor  will,  as  we  have  seen, 
occupy  such  a  position  that  if  the  line  CGt  be  drawn  at  the  angle, 
X  to  00l}  the  zero  of  the  current  band  will  be  along  CGl  and  its 
maximum  along  DD±  at  right  angles  thereto.  Consider  any  point 
P  on  the  circumference  making  the  angle  6  with  OOj.  Then  the 
flux  in  the  air  gap  at  that  point  will  be  4>  sin  0,  while  the  current 
there  will  be  C2sin(0  —  X).  Consider  the  arc  of  the  rotor  sub- 
tended by  the  small  angle  dd,  in  that  portion  of  the  rotor  the  total 
current  will  be  p  .  C2  sin  (6  —  X)  dQ,  and  thus  the  torque  on  that 
part  will  be  given  by  the  product  of  the  flux,  the  current  and  the 
radius,  or  will  be 


The  torque  produced  by  the  whole  armature  will  evidently  then 
be  given  by 

P  =  p2 .  cj> .  C2 .  I  'sin  0 .  sin  (6  -  X)  dB, 

Jo 

on  evaluating  the  integral  this  gives  P  =  pz .  4> .  C2 .  TT  .  cos  X,  but 
we  see  from  Fig.  153  that 

Or  r 

cosX  =  7-^r=  , 

C/£2     Vr2  +  ^Vv 


184 

11 
and  we  also  have 


ALTERNATING   CURRENTS 


<I>.p.cr 

-- 


Vr2 


Hence  the  final  expression  for  the  torque  is 


rer 


For  the  immediate  purpose  in  hand  it  is  enough  to  consider  that 
the  air  gap  flux  is  constant  under  all  circumstances,  in  which  case 
we  can  put 

P-k 

^~ 


In  terms  of  the  total  flux  <£>,,  crossing  the  air  gap  we  can  write 


7T 


k  =  -T  .  p  .  <£>/.     For  the  mean  air  gap  flux  is  evidently  -  <I>,  and 
hence  the  total  flux  is 


7T 


X  7T/3 


2  . 


It  should  be  noted  that,  other  things  being  equal,  the  torque 
is  proportional  to  4>2.  Now  the  induced  pressure  in  the  stator  at 
constant  periodicity  is  proportional  to  <3>,  and  since  we  have  seen 
that  the  impressed  pressure  is  roughly  equal  to  the  induced  one, 
we  can  assume  that  the  torque  will  be  approximately  proportional 
to  the  square  of  the  impressed  pressure. 

Running  condition.  Some  special  conditions  of  operation 
must  now  be  considered.  Take  the  case  of  running  under  load ; 
since  the  loss  in  the  rotor  is  proportional  to  the  slip,  a,  it  will  be 
small  under  such  conditions  and  the  diagram  (Fig.  155)  will  represent 
the  state  of  things.  When  there  is  no  load  externally  applied  to 


THE   INDUCTION   MOTOR 


185 


the  rotor  the  slip  will  be  extremely  small  and  the  current  and 
leakage  in  the  rotor  practically  zero,  thus  the  angle  of  lag  in  the 
rotor  wires  will  be  nearly  zero  and  the  diagram  will  be  somewhat 
as  shown  in  Fig.  156.  Owing  to  the  large  air  gap  which,  as 
we  have  seen,  necessitates  a  comparatively  large  magnetising 
current,  there  will,  even  in  this  case,  be  a  fair  amount  of  leakage 


Fig.  155. 


Fig.  156. 


flux  in  the  stator,  and  hence  the  corresponding  E.M.F.  will  have 
considerable  value,  thus  at  no  load  there  will  be  quite  a  large 
angle  of  lag,  yfr,  in  the  stator  current. 

Starting.  The  conditions  at  starting  involve  the  slip  being  fl 
and  thus  cause  both  the  current  in  the  rotor  and  the  angle  of  lag 
between  this  and  the  induced  pressure  to  be  very  great,  the 
leakage  E.M.F.  in  the  rotor  being  excessive  owing  to  the  high  slip. 
This  means  that  the  current  in  the  stator  must  also  be  very  large, 
the  leakage  flux  thereof  also  very  large,  and  consequently  the 
phase  angle  between  the  current  and  pressure  also  very  large. 
Fig.  157  will  show  the  relations  existing  in  this  case. 

It  will  be  seen  that  if  the  flux  in  the  air  gap  be  constant  as  we 
have  assumed,  the  impressed  pressure  required  is  much  greater  than 
in  the  previous  cases.  But  the  quantity  that  is  actually  kept  con- 
stant being  (as  in  the  transformer)  this  pressure,  it  follows  that  the 
flux  will  be  largely  reduced  at  starting,  and  hence  the  torque  will 


186 


ALTERNATING   CURRENTS 


be  much  smaller  than  that  corresponding  to  the  assumption  of  con- 
stant air-gap  flux.     Furthermore,  the  current  demanded  from  the 


Fig.  157. 

source  of  supply  is  very  large,  many  times  the  maximum  value 
that  is  wanted  in  ordinary  running,  and  hence  some  method  must 
be  found  to  better  the  starting  conditions.  We  shall  describe  in 


E/.E2 


Fig.  158. 


THE   INDUCTION   MOTOR 


187 


Chap.  XVII  methods  that  can  be  used,  but  just  now  we  shall  show 
that  if  by  any  means  we  can  increase  the  resistance  of  the  rotor 
wires  at  starting  a  much  better  state  of  things  will  be  available,  the 
method  of  doing  this  will  be  seen  later  on.  In  this  case  with  the 
maximum  slip  of  O  the  E.M.F.  in  the  rotor  wires  will  be  the  same 
as  before,  but  since  the  resistance  of  the  wires  is  increased,  the 
current  and  the  angle  of  lag  will  be  much  diminished.  Thus  the 
stator  current  will  also  be  diminished  together  with  its  leakage 
field  and  the  angle  of  lag  as  shown  in  Fig.  158.  It  will  follow  that 
the  torque  will  be  actually  increased  at  starting.  That  this  is  the 
case  when  we  assume  constant  air-gap  flux  is  evident  from  the 
expression  for  the  torque.  For  the  value  of  LSI  is  then  big 
compared  with  r  and  thus  the  torque  at  starting  is  nearly  given 

r£l        r 
by  Pg  =  y—  -  =  y^r,  that  is,  it  is  proportional  to  the  rotor  resistance. 


Mechanical  characteristic.  Under  the  assumption  that 
the  air-gap  flux  is  constant  for  all  loads  on  the  motor,  which  is  only 
absolutely  true  when  the  slip  is  as  small  as  it  usually  is  in  nearly  all 
practical  conditions  of  running,  the  expression  for  the  torque  as  a 
function  of  the  slip  enables  us  to  derive  the  relation  between  the 
torque  and  the  angular  velocity  of  the  motor,  or  its  mechanical 
characteristic.  Take  the  line  OX  (Fig.  159)  with  a  length  equal  to 
the  angular  velocity  (O)  of  the  rotating  field  on  any  assumed 
scale,  and  let  a  scale  of  torques  be  drawn  along  OT.  With  X  as 
origin  draw  the  trace  of  the  relation  given  on  p.  184  between  the 


OS  N 

Torque 

Fig.  159. 

torque,  P,  and  slip,  <T.  If  P  be  any  point  on  the  curve  thus 
obtained  we  have  PM  equal  to  the  slip  and  MN  to  the  constant 
angular  velocity  of  the  field  or  H.  But  if  co  is  the  angular 
velocity  of  the  rotor  we  also  have  fl  =  co  +  a,  and  thus  PN  is  this 


188  ALTERNATING   CURRENTS 

angular  velocity.  It  follows  that  the  curve  is  also  the  mechanical 
characteristic  of  the  motor,  that  is,  the  relation  between  torque 
and  speed,  provided  OX  is  taken  as  the  axis  of  angular  velocity. 
It  will  be  seen  that  the  torque  attains  a  maximum  value  at  the 
point  Q,  from  X  to  Q  the  torque  increases  with  fall  in  speed ;  that 
part  of  the  curve  is  the  only  actually  existent  portion  of  the 
mechanical  characteristic:  from  Q  to  8  the  opposite  condition 
prevails  and  the  relation  is  unstable  with  ordinary  forms  of  brake 
or  load.  The  abscissa  OS  is  the  torque  that  is  produced  when  the 
rotor  is  at  rest  and  consequently  represents  the  starting  value  of 
the  torque.  If  the  opposing  torque  at  that  moment,  whether 
internal  or  total,  exceeds  that  amount,  the  rotor  will  not  start. 
If  less,  it  runs  up  to  the  proper  speed  corresponding  to  the  torque 
required  as  given  by  the  stable  part  of  the  curve,  XQ. 

Consider  the  part  of  the  curve  extending  nearly  up  to  the 
point  of  maximum  torque.  In  any  practical  rotor  which  is  re- 
quired to  work  constantly  it  is  necessary,  for  reasons  of  efficiency, 
to  keep  the  slip  small,  and  in  this  case  La-  is  small  compared  with 

r.     Thus  the  running  torque  is  nearly  given  by  Pr  —  k.-  or  Pr 

is  proportional  nearly  to  the  slip.  Hence  the  greater  portion  of  XQ 
is  nearly  a  straight  line.  It  follows  that  such  a  motor  as  we  are 
considering  will  run  from  zero  torque  to  nearly  its  maximum 
possible  torque  with  approximately  constant  speed,  and  is  thus 
very  like  the  direct  current  shunt  motor  in  its  mechanical  pro- 
perties. Furthermore  it  is  seen  that  Pr  is  inversely  as  the  rotor 
resistance,  r,  and  hence  for  a  definite  slip  the  torque  will  be 
greater  the  smaller  r  is.  Hence  it  is  desirable  in  general  to  keep 
the  rotor  resistance  as  low  as  possible.  In  some  cases,  such  as 
motors  for  cranes,  the  important  thing  is  not  constancy  of  speed  or 
even  high  efficiency,  but  the  production  of  high  starting  torque, 
and  in  such  a  case  the  short-circuiting  ring  of  the  rotor's  rods  is 
made  of  some  metal  of  higher  specific  resistance  than  copper. 

Dynamo  action.  We  might  enquire  what  happens  for 
angular  velocities  outside  the  ranges  in  Fig.  159.  Thus  if  the 
rotor  be  not  merely  at  rest  but  be  driven  in  such  a  way  that  it  is 
rotating  in  the  opposite  direction  to  the  field,  we  get  the  continua- 
tion SB  of  the  curve  (Fig.  160).  Since  the  rotation  is  in  the 
opposite  direction  to  the  couple  that  the  rotor  produces,  the 
machine  will  be  a  generator  and  not  a  motor.  Again,  let  the 
rotor  be  driven  above  the  synchronous  speed,  &,  the  slip,  being 
still  given  by  the  relation  a-  =  £1  —  o>,  will  be  negative  since  o>  is 
now  greater  than  H.  Thus  the  torque  speed  curve  will  be  as  in 
the  part  XF  of  the  curve  having  the  same  form  as  the  original 
part,  but  on  the  opposite  side  of  the  axis  OX.  In  this  case  the 
angular  velocity  is  in  the  same  direction  as  for  the  motor,  but  the 
torque  is  opposite  in  sign,  hence  the  machine  is  again  acting  as  a 
generator,  that  is  if  the  stator  be  connected  to  mains  in  which  the 


THE    INDUCTION    MOTOR 


189 


proper  phase  relation  between  the  pressures  is  maintained,  and  if 
the  rotor  of  the  machine  is  driven  by  external  means  so  as  to  run 
above  the  synchronous  speed,  power  will  be  delivered  to  the  mains 


Fig.  161. 


190 


ALTERNATING   CURRENTS 


attached  to  the  stator.  Such  a  machine  is  called  an  asynchronous 
generator.  It  must  be  noted  that  it  cannot  of  itself  produce 
power,  the  stator  must  be  excited  by  the  properly  phased  currents. 
One  important  property  of  such  a  machine  is  that  it  evidently 
delivers  a  current  into  the  mains  which  leads  the  pressure  instead 
of  lagging  behind  it ;  it  therefore  tends  to  improve  the  power  factor 
of  the  system. 

Actual  form  of  motor.  Fig.  161  shows  a  complete  in- 
duction motor  of  the  form  we  have  been  considering,  while 
Figs.  162  and  163  show  the  construction  of  the  stator  and  rotor  of 
the  same.  The  form  of  winding  of  the  stator  can  readily  be  seen 
as  well  as  the  method  in  which  the  short-circuiting  of  the  rotor 
bars  is  carried  out. 


Fig.  162. 


Fig.  163. 


CHAPTER  XV. 

THE   HEYLAND   CIRCLES. 

THE  following  very  elegant  construction  for  investigating  the 
relations  between  the  various  quantities  in  a  rotary  field  motor 
has  been  given  by  Mr  Heyland,  and  is  generally  known  as  the 
Heyland  diagram.  Since  the  applied  pressure,  £0 ,  is  constant,  the 
maximum  stator  flux  must  likewise  be  a  constant,  let  it  be  denoted 
by  <l>.  Let  QS,  Fig.  164,  be  this  constant  stator  flux  and  QR  the 
nett  rotor  flux  as  in  the  previous  figure  (Fig.  153).  Then  SR  will 
be  the  total  leakage  flux  between  the  rotor  and  stator.  From  the 
same  figure  it  will  be  seen  that  the  two  leakage  fluxes  from 
the  stator  and  the  rotor  are  very  nearly  in  a  line,  and  in  most 
cases  little  error  will  be  made  if  it  be  taken  that  all  the  leakage 
flux  between  the  two  is  in  phase  with  the  stator  current  and 
proportional  thereto.  Redraw  this  triangle  as  at  OPD,  the  sides 
of  the  latter  triangle  being  perpendicular  to  those  of  the  first, 
and  the  arrows  on  the  sides  showing  the  actual  direction  of  the 
different  fluxes.  Then  since  the  leakage  flux  has  been  taken  as 
proportional  to  the  stator  current  the  line  OP  can  be  taken  to 
represent  the  value  of  the  stator  current  as  well  as  that  of  its 
leakage  flux.  Since  OD  represents  the  total  constant  stator  flux 
3>,  it  will  be  of  fixed  length,  as  the  back  E.M.F.  due  to  it  has  to  be 
equilibrated  by  the  impressed  pressure.  Furthermore  DP  is  the 
nett  rotor  flux,  and  from  Fig.  153  it  will  be  seen  that  the  rotor 
current  is  in  phase  with  that  flux.  Hence  if  a  line  PG  be  drawn 
perpendicular  to  DP  (that  is,  parallel  to  QR  the  rotor  flux),  it 
will  give  the  actual  direction  of  the  rotor  current. 

The  rotor  flux  is  due  to  the  combined  action  of  the  stator  and 
rotor  bands,  and  hence  the  band  of  current  for  that  flux  must 
be  given  by  the  resultant  of  the  vectors  representing  those  bands, 
but  it  must  also  fulfil  the  proper  space  relation  between  a  flux  and 
its  corresponding  magnetising  band,  that  is  to  say,  it  must  be 
in  space  at  an  angle  of  90°  to  its  flux,  hence  the  direction  of  this 
resultant  of  the  stator  and  rotor  bands  of  current  must  be  in  the 
direction  parallel  to  DP,  that  is,  perpendicular  to  the  direction  of  the 
rotor  flux  RQ.  Let  PG  cut  OD  in  C,  and  on  OCD  draw  the  semi- 


192 


ALTERNATING  CURRENTS 


circles  shown.  Draw  OG  parallel  to  PD;  since  OP  gives  the  value 
of  the  stator  current,  and  its  two  components  (the  nett  rotor  current 
and  the  magnetising  current)  have  respectively  the  directions  given 
by  PG  and  OG,  it  follows  that  these  latter  quantities  must  be 
represented  by  those  vectors. 


Since  the  nett  rotor  flux  is  due  to  the  mi 
OG,  and  since  the  rotor's  reluctance,  p,  has  been  taken  as  constant, 
it  follows  that  DP  and  OG  have  a  constant  ratio.  Further,  the 
angles  DPC  and  CGO  are  necessarily  both  right  angles,  hence  it 
follows  that  the  curves  on  which  the  points  P  and  G  move  will  be 
two  fixed  semicircles.  It  also  follows  that  the  line  OD  is  divided 
in  some  definite  ratio  at  C,  and  this  ratio  must  now  be  found. 
Suppose  the  machine  to  be  running  in  such  a  condition  that  there 
is  no  leakage  whatever,  all  the  constant  flux,  <£,  passes  directly 
into  the  rotor  and  thus  only  encountering  the  corresponding 
reluctance  p.  This  could  be  nearly  realised  if  we  imagine  the 
rotor  to  be  running  in  absolute  synchronism  with  the  field.  Then 
the  only  current  band  existing  would  evidently  be  given  by  OG,  as 
P  will  have  come  down  to  the  point  C.  Hence  if  k  be  some 
definite  constant  depending  solely  and  entirely  on  the  form  and 
amount  of  the  stator  windings,  the  magnetomotive  force  acting 
will  be  given  by  k .  OC,  and  this  will  then  succeed  in  forcing  the 
definite  flux  4>  through  the  reluctance  p.  We  thus  have 


THE   HEYLAND   CIRCLES  193 

k  .  OC=  0/3.  Now  let  the  same  definite  flux  be  imagined  to  entirely 
pass  through  the  stator  leakage  paths  the  reluctance  of  which  is 
pl  ;  this  means  that  P  must  move  down  to  D  and  that  the  current 
required  in  the  stator  will  be  given  by  OD,  the  rotor  current  being 
exactly  the  same  in  amount  and  so  opposing  the  stator  band  that 
no  flux  can  enter  the  rotor.  Since  the  current  given  by  OD 
is  passing  through  the  same  circuit  as  before,  the  magnetomotive 
force  produced  will  be  k  .  OD,  and  since  the  resultant  flux  produced 
is  taken  to  be  the  definite  amount  <t>,  represented  by  QS,  we  now 
have  k  .  OD  =  4>pj  .  These  two  equations  lead  to  OD/OC  =  pjp  =  v, 
and  hence  the  line  OD  is  divided  in  C  in  such  a  way  that 


also  DC:OD::(v-I):v,   or   DC  =  z.OD, 

where  a  is  a  constant. 

It  is  evident  that  it  is  not  necessary  to  draw  the  little 
semicircle  in  order  to  get  the  rotor's  current,  for  we  have 
PC/PG  =  DC/  DO  or  the  rotor's  current  given  by  PG  is  also  given 


Torque  and  slip  lines,  ideal  case.  Assume  that  no  losses 
occur  in  the  machine,  then  it  can  easily  be  shown  that  other  im- 
portant quantities  can  be  represented  by  lines  on  this  diagram  ;  thus 
if  the  line  PT  be  drawn  perpendicular  to  OD  from  P  we  can  show 
that  this  line  is  proportional  to  the  torque  exerted  by  the  motor. 
For  the  torque  is  proportional  to  the  product  of  the  current  in  the 
rotor  multiplied  by  the  flux  that  the  rotor  is  cutting  at  that 
instant.  Now  the  current  is  given  by  PG  and  the  corresponding 
flux  in  phase  with  it  in  space  by  the  line  QR.  Hence  the  torque 
is  proportional  to  the  product  PG  .  QR.  But  we  made  DP 
proportional  to  QR  and  we  know  that  PG  is  proportional  to  PC, 
hence  the  torque  is  proportional  to  the  product  DP.  PC.  But 
from  the  similar  triangles  DPC  and  DTP  we  have 

PC/CD  =  PT/DP  or   DP  .  PC  =  PT.  CD, 

but  since  CD  is  constant  the  torque  is  proportional  to  the  line  PT. 
The  maximum  torque  will  then,  we  see,  occur  when  the  point  P  is 
at  the  top  of  the  semicircle. 

Again,  draw  a  perpendicular  from  C  and  produce  DP  to  cut  it 
in  W.  The  line  C  W  is  proportional  to  the  slip.  For  we  see  from 
the  considerations  on  p.  175,  that  the  current  in  the  rotor 
is  always  just  proportional  to  the  rate  at  which  the  flux  in  phase 
with  that  current  is  cut,  and  hence  since  the  current  in  the  rotor 
is  proportional  to  PG  and  the  field  to  QR,  the  slip  is  proportional 
to  PG/QR  which  is  as  before  proportional  to  PC/PD.  But  in  the 

L.  13 


194  ALTERNATING   CURRENTS 

similar  triangles  CPD  and  DCW  we  have  PC/PD  =  CW/CD  and 
the  line  CD  is  constant,  hence  a  is  proportional  to  C.W. 

The  maximum  slip  will  occur  at  standstill  of  the  motor, 
corresponding  to  a  having  the  value  H,  and  hence  the  line  C  W  has 
a  finite  length ;  it  follows  that  the  position  for  the  current  vector 
OP  at  standstill  is  at  some  such  point  as  P^  not  extending  down 
to  D.  Again,  when  running  light  there  is  still  a  definite,  though 
small,  torque  required  by  the  rotor  owing  to  frictional  and 
other  losses,  and  hence  the  current  vector  OP  will  have  a  similar 
limiting  position  P2  on  the  other  side  of  the  circle. 

Applied  pressure  line.  The  position  of  the  applied  pressure 
vector  can  be  found  thus.  Since  QS  is  the  total  flux  through  the 
stator  the  corresponding  total  induced  E.M.F.  in  it  will  be  in  phase 
with  the  flux  QS,  hence  the  impressed  pressure  will  be  in  antiphase 
with  that  flux,  and  will  be  represented  by  the  line  0  V  drawn  from 
0  equal  to  the  induced  primary  pressure  but  in  the  opposite 
direction.  Thus  the  angle  of  lag,  X,  will  be  POFand  the  power 
factor  will  be  the  cosine  of  that  angle.  But  since  the  angle  POC 
is  the  complement  of  POV  the  power  factor  is  also  the  sine  of 
POO  or  is  given  by  CZ/OC  where  CZ  is  the  perpendicular  from  C 
on  the  current  vector,  but  since  OC  is  constant  the  power  factor  is 
proportional  to  the  line  CZ.  It  is  evident  that  this  has  its  maxi- 
mum value  when  the  line  OP  touches  the  circle,  and  hence  the 
corresponding  current  should  be  the  one  at  which  the  motor  does 
most  of  its  work. 

Application  to  actual  motor.  The  diagram  thus  derived 
refers  to  the  relative  values  of  the  maxima  and  angular  positions 
of  the  different  quantities  concerned  in  the  case  of  the  idealised 
machine,  and  for  this  aspect  of  the  question  the  whole  diagram 
must,  as  has  been  said,  be  thought  of  as  rotating  with  constant 
angular  velocity  equal  to  the  periods  per  second  of  the  applied 
pressure.  Now  consider  a  single  phase  of  the  machine,  and  let 
the  projections  of  the  several  current  and  pressure  vectors  on  any 
line  be  taken.  It  is  evident  that  these  projections  will  give  the 
corresponding  instantaneous  values,  both  in  value  and  phase 
relation,  of  the  corresponding  quantities  per  phase,  that  is  of  the 
alternating  bands  of  current,  etc.  in  any  one  of  the  windings  of  our 
idealised  machine.  Depending  on  the  connections  of  the  winding, 
whether  two-  or  three-phase,  so  must  we  take  two  or  three  lines  of 
reference  at  the  proper  relative  angles  to  get  a  full  representation 
of  the  events  in  the  separate  circuits.  Now  we  must  consider 
that  each  circuit  executes  the  same  cycle  as  the  others,  and  hence 
any  one  can  be  taken  as  representative  of  all,  the  only  difference 
between  the  successive  circuits  being  that  the  events  occur  either 
\  or  £  of  a  period  later  in  time.  This  being  so,  when  we  are 
considering  the  same  quantities  for  any  phase  of  any  motor,  it  is 


THE   HEYLAND   CIRCLES  195 

evident  that  instead  of  taking  the  vectors  as  representing  the 
maximum  value  of  the  quantities  concerned  with  one  of  our 
imaginary  circuits,  we  can  equally  well  take  them  to  represent  the 
maximum  values,  both  in  magnitude  and  phase,  of  the  corresponding 
quantities  referred  to  the  actual  circuit  of  the  machine.  Further, 
since,  with  the  assumed  sinusoidal  law  of  pressures,  etc.,  the 
maximum  bears  a  definite  ratio  to  the  virtual  value,  we  can  use 
directly  the  virtual  values  of  the  different  quantities  (current  and 
pressure)  without  any  loss  of  generality.  It  follows  that  a  similar 
construction  is  available  for  any  induction  motor  where  the 
quantities  involved  are  no  longer  the  currents,  fluxes,  and  pressures 
in  the  idealised  equivalent  winding,  but  the  actually  existent 
measured  ones  corresponding  to  any  phase  of  the  actual  stator  of 
the  motor. 

Motor  with  losses.  The  diagram  developed  on  p.  190  refers 
to  a  motor  in  which  all  the  losses  were  neglected,  nothing  having 
so  far  been  said  as  to  these  quantities.  It  is  now  necessary  to 
see  if  it  is  possible  to  modify  the  diagram  in  such  a  manner 
that  it  will  give  a  suitable  representation  of  the  actual  condition 
of  affairs  in  the  motor  when  these  losses  are  taken  into  account. 
The  losses  fall  broadly  into  two  categories,  those  due  to  purely 
ohmic  resistance  in  the  windings  of  the  stator  and  rotor,  and  those 
incidental  to  the  rotation  of  the  latter  in  the  air-gap  field. 

Representation  of  ohmic  losses.  Take  any  line  0V 
(Fig.  165)  to  give  the  direction  of  the  applied  pressure  on  any  phase 
of  the  stator,  and  for  the  present  let  it  be  assumed  that  we  have 
drawn  on  the  line  OD  perpendicular  to  OF  a  Heyland  circle 
OMD  as  before :  the  method  of  actually  obtaining  this  circle  will 
be  given  later  on.  We  will,  in  accordance  with  what  has  just  been 
said,  take  it  that  this  refers  to  any  phase  of  the  stator,  that  is,  if  P 
is  any  point  on  this  circle,  OP  will  represent  the  virtual  stator 
current  in  magnitude  and  the  phase  of  its  maximum  relative  to  0  V. 
Hence  from  what  we  have  done  before  it  will  follow  that  DP 
represents  the  rotor  field  in  phase  and  magnitude,  assuming  that 
resistance  is  absent. 

The  effect  of  resistance  in  the  stator  can  be  represented  as 
follows:  the  actual  stator  current  OP  is  equivalent  to  the  two 
currents  OC  and  PC,  of  which  the  former  is  constant  and  the  latter 
increases  with  the  load  on  the  motor.  Hence  any  loss  of  energy 
due  to  the  passage  of  the  current  OC  falls  into  the  category 
of  losses  incident  to  the  production  of  the  flux,  and  can  thus 
be  left  for  future  consideration.  Hence  the  drop  of  pressure 
incident  to  varying  load  is  proportional  to  PC.  The  condition  of 
operation  of  the  motor  is  that  the  applied  pressure  is  constant, 
and  thus  it  follows  that  any  diminution  of  pressure  due  to  drop 
in  the  stator  wires  must  be  subtracted  vectorially  from  the  applied 

13—2 


196 


ALTERNATING   CURRENTS 


pressure  in  order  to  obtain  the  residue  that  has  to  equilibrate 
the  induced  pressures.  Thus  both  the  induced  pressure  due  to 
the  leakage  field  and  that  due  to  the  rotor  flux  will  be 
necessarily  less  when  a  drop  occurs  in  the  stator  due  to  its 
resistance.  It  follows  that  the  flux  is  diminished  owing  to  this 
drop,  and  it  can  be  taken  that  it  is  diminished  in  proportion 
to  the  drop  in  the  stator. 


Fig.  165. 

Thus  let  a  point  Q  be  taken  on  DP  such  that  PQ  bears 
a  constant  ratio  to  the  variable  part  CP  of  the  stator  current ;  the 
effect  of  stator  drop  can  then  be  represented  by  assuming  that  the 
rotor  flux  is  reduced  from  DP,  the  value  it  would  have  with  no 
drop,  to  DQ.  Now  since  the  angle  DPC  is  a  right  angle,  and  the 
sides  CP,  PQ  of  the  triangle  CPQ  are  always  in  a  constant  ratio, 
this  triangle  must  remain  of  the  same  shape  for  all  positions 
of  the  point  P,  and  hence  the  angle  DQC  has  always  the  same 
value.  Thus  the  locus  of  Q  is  the  circle  CQD  as  shown. 

The  effect  of  drop  in  the  rotor  can  be  represented  in  exactly 
the  same  way.  For  CQ  is  also  proportional  to  CP,  and  hence  to  the 
rotor  current  as  we  have  seen,  hence  if  the  point  R  be  taken  on 
DP  such  that  RQ  bears  a  constant  ratio  to  CQ,  the  line  DR  will 
represent  the  value  of  the  nett  rotor  flux,  after  all  the  drops  of 
pressure  are  allowed  for.  By  similar  reasoning  to  the  last  case, 
the  point  R  will  describe  the  circle  shown.  Hence  the  effect  of 
the  two  drops  in  rotor  and  stator  can  be  represented  by  drawing 
in  the  two  extra  circles. 

Representation    of    core    and    rotational    losses.       It 

remains  now  to  consider  the  other  sources  of  loss.     As  far  as  the 


THE   HEYLAND   CIRCLES  197 

stator  is  concerned  these  are,  hysteresis,  eddy  currents,  and  the 
constant  ohmic  loss  due  to  the  current  represented  by  OC.  The 
periodicity  of  the  applied  flux  in  any  phase  of  the  stator  is 
constant,  and  the  value  of  that  flux  will  diminish  slightly  with 
increase  of  load.  As  regards  the  rotor  the  principal  loss  is  the 
constant  frictional  one;  as  regards  the  core  losses,  when  the 
slip  is  small  the  flux  is  large,  and  when  the  slip  is  very  great 
the  induced  rotor  currents  force  the  flux  out,  and  hence  the 
flux  is  small.  A  very  close  approximation  to  the  truth  is 
then  arrived  at,  if  we  assume  that  the  total  loss  of  energy- 
incident  to  the  magnetisation  and  rotation  of  the  motor  is 
nearly  a  constant  quantity.  This  can  be  represented  in  the 
diagram  as  follows.  Draw  a  scale  of  current  appropriate  to  that 
used  for  the  diagram  along  the  line  OF,  and  set  off  a  point  I  such 
that  it  represents  a  current  that,  flowing  in  phase  with  the  given 
applied  pressure,  will  represent  this  constant  loss  of  energy,  and 
draw  a  line  II  parallel  to  DO. 

Input,  output,  torque,  and  slip  lines.  We  will  now 
see  how  we  can  represent  the  other  related  quantities  in  the 
amended  diagram.  Firstly,  as  regards  the  input :  the  current 
flowing  into  any  phase  of  the  stator  is  given  by  OP,  and  hence  the 
part  of  this  current  that  is  in  phase  with  the  terminal  pressure 
will  represent  to  some  scale  the  input.  If  Pp  is  drawn  from 
P  perpendicular  to  OD  it  will,  therefore,  be  proportional  to  the 
input. 

Secondly,  to  find  a  line  giving  the  torque :  the  nett  air-gap  flux 
is  given  by  DQ  and  the  rotor  current  is  proportional  to  CP,  hence 
the  total  torque  produced  is  proportional  to  DQ  x  CP,  that 
is  to  the  area  of  the  triangle  DQG,  since  DQ  is  its  base  and  CP  is 
its  height.  But  this  triangle's  area  is  also  given  by  Qql  x  CD,  and 
the  line  CD  is  a  fixed  one,  hence  the  total  torque  produced  is 
proportional  to  Q</I.  But  the  incidental  loss  in  the  rotor  is 
evidently  equivalent  to  a  torque  proportional  to  qqlt  and  hence  the 
nett  available  torque  is  given  by  Qq. 

Thirdly,  to  find  a  line  giving  the  output :  this  is  proportional 
to  the  nett  pressure  induced  in  the  rotor  multiplied  into  the 
rotor's  current,  but  the  pressure  produced  is  proportional  to  the 
nett  field  acting,  that  is  to  DR :  hence  the  output  will  be  given  by 
DR  x  CP  and  hence  by  Ri\  x  CD.  As  before  (when  the  loss  due 
to  the  constant  effects  is  considered),  instead  of  R^  giving  the 
output,  it  will  be  proportional  to  Rr. 

Fourthly,  to  find  a  line  representing  the  slip :  this  is  evidently 
proportional  to  the  rotor  current  divided  by  the  air-gap  flux,  or  is 
PC/DQ.  But  PC  is  proportional  to  CQ,  hence  the  slip  is  given 
by  CQ/DQ.  Draw  any  line  such  as  MS  (the  reason  for  the 
selection  of  the  special  point  M  will  be  given  later  on),  in  such  a 


198  ALTERNATING   CURRENTS 

way  that  the  angle  MsD  is  equal  to  the  angle  DQC  ';  this  can 
readily  be  done  by  means  of  a  piece  of  tracing-paper.  Then  the 
triangles  DSs  and  DQC  are  similar,  and  hence  the  slip  is  pro- 
portional to  Ss/Ds,  or  since  Ds  is  a  fixed  length,  the  slip  is  given 
by  the  distance  Ss. 

Experimental  determination  of  the  circles.  It  follows 
that,  provided  we  can  in  some  manner  obtain  the  actual  circles  for 
any  motor,  it  is  easy  to  draw  up  tables  of  the  various  quantities 
involved  which  will  enable  the  performance  of  the  motor  to  be 
predicted.  We  must  now  see  how  the  diagram  can  be  found  for 
an  actual  motor.  The  line  0  V  being  taken  and  a  line  of  indefinite 
length  at  right  angles  thereto,  if  in  any  way  we  can  find  the 
coordinates  of  any  two  points  on  the  outer  circle  (since  its  centre 
must  be  on  the  perpendicular  to  0  V)  that  circle  will  be  completely 
determined.  These  two  points  can  be  determined  as  follows, 
where  for  simplicity  of  explanation  actual  figures  pertaining  to  a 
definite  motor  are  employed.  The  motor  was  first  run  without  any 
load,  which  is  called  the  No-load  test,  and  the  total  power  taken, 
the  current  per  main  and  the  pressure  of  supply  measured.  In 
the  case  considered,  the  motor  was  four  pole  running  at  1300 
R.P.M.  synchronous  speed  with  43  periods,  it  had  a  three-phase 
stator  wound  in  the  star  manner,  and  the  pressure  between  the 
mains  was  120  volts,  the  current  in  one  main  4*1  amperes,  and  the 
total  watts  210.  Assuming  that  we  can  treat  the  pressures  as 
sinusoidal,  the  pressure  at  the  terminals  of  one  of  the  legs  of  the 
star  was  120/V3  :  the  watts  taken  by  one  of  the  phases  was  70, 
and  hence  it  follows  that  the  current  in  phase  with  the  pressure 

was  or  1  ampere.     The  wattless  current,  that  in  quadrature 


with  the  pressure,  is  then  given  by  Vl4'l2  —  I2  or  4  amperes. 
Hence  set  out  the  point  L  so  that  the  distance  01  is  one  ampere 
and  the  distance  OC  is  four  amperes.  This  gives  the  position 
of  one  point  on  the  circle.  It  may  be  noted  that  the  constant  loss 
line  can  at  once  be  drawn  through  the  point  /. 

To  determine  another  point,  let  the  armature  be  blocked 
so  that  it  cannot  rotate,  and  again  measure  the  current  taken 
per  phase  and  the  total  watts,  the  pressure  across  the  mains  being 
maintained  at  its  former  value.  This  is  called  the  Stand-still  test. 
In  this  case  the  current  was  55  amperes,  and  the  power  6940 

6940 
watts.     It    follows   that   the  watts   per  phase   are  —  ^—  and   the 

current  in  phase  with  the  pressure  is  ^  —  z™  or  33*5  amperes  ; 

o  X 


hence  the  wattless  component  of  the  current  is  \/552  —  33'52 
or  43'5  amperes.  Set  off  the  distance  ON  equal  to  33'5  on  the 
current  scale,  and  the  distance  MN  equal  to  43*5,  the  point  M  is  a. 


THE   HEYLAND   CIRCLES  199 

second  point  on  the  circle.  The  centre  can  then  be  found  by 
joining  LM,  bisecting  it  in  U  and  drawing  the  perpendicular  UX 
to  cut  the  indefinite  perpendicular  line  from  0  in  the  point  X. 
A  circle  drawn  with  X  as  centre  is  the  outer  one  of  our  three 
circles. 

In  many  motors  the  current  that  would  flow  in  the  stator 
under  the  standstill  condition  would  be  much  larger  than  could 
be  safely  permitted.  In  such  a  case  a  pressure  less  than  the 
normal  value  must  be  used  for  the  test  in  order  to  keep  the 
current  down  to  a  reasonable  value.  Having  determined  this 
reduced  pressure,  the  total  power  taken  and  the  current,  the  two 
components  of  the  current  should  be  calculated  under  the  con- 
ditions of  the  test.  To  find  what  they  would  have  been  if  the 
pressure  employed  had  had  the  proper  normal  value,  it  is  only 
necessary  to  increase  each  in  the  ratio  that  the  normal  pressure 
bears  to  the  pressure  actually  employed  in  the  test.  This  will 
evidently  follow  from  the  fact  that  the  leakage  fluxes  are  pro- 
portional to  the  current  only. 

The  inner  circle  can  be  very  simply  found:  OM  gives  the 
current  at  standstill,  and  at  that  point  the  total  output  is 
evidently  zero  since  the  rotor  is  at  rest,  hence  when  P  moves 
up  to  Mt  R  moves  on  to  D,  from  which  it  follows  that  MD  is  a 
tangent  to  the  inner  of  the  circles,  hence  if  a  perpendicular  be 
drawn  from  X  and  be  cut  in  Z  by  the  normal  to  MD,  Z  is  the  centre 
of  the  inner  circle. 

The  middle  circle  must  now  be  determined,  which  can  be  done 
as  follows:  at  standstill  all  the  nett  flux  left  is  used  up  in 
providing  the  necessary  pressure  to  force  the  stator  current 
through  its  resistance  and  the  rotor  current  through  its  resistance. 
Hence  the  flux  given  by  DM  has  to  fulfil  these  conditions  only. 
At  standstill  the  flux  corresponding  to  the  pressure  required 
to  force  the  current  through  the  stator  will,  from  the  construction, 
be  given  by  MT  (since  it  is  represented  by  PQ  in  the  general 
consideration,  when  the  current  is  given  by  OP),  the  value  of  that 
pressure  will  be  the  product  of  the  standstill  current  and  the 
stator  resistance  per  phase.  In  the  present  case  that  resistance 
was  0'344  ohm,  so  that  the  pressure  under  consideration  must  be 
18*9  volts.  It  remains  for -us  to  determine  how  this  can  be 
represented  on  the  corresponding  flux  vector  MT.  It  must  be 
noted  that  the  line  OD  represents  the  constant  flux  corresponding 
to  the  constant  applied  pressure  of  120/\/3  volts  that  exists  on  the 
terminals  of  one  phase  of  the  stator,  hence  if  the  flux  vector  OD 
represents  that  flux  which  is  necessary  to  produce  120/\/3  volts, 
the  voltage  corresponding  to  the  vector  DM  will  be DM  I OD  times 
that  amount.  On  scaling  off  the  lengths  of  the  vectors  OD  and 
DM,  and  multiplying  this  ratio  by  the  applied  stator  pressure,  we 
find  that  the  pressure  corresponding  to  DM  is  41 '2  volts,  hence  to 


200 


ALTERNATING   CURRENTS 


determine  the  position  of  the  point  T  we  must  divide  DM  at  T 
into  two  parts  such  that  MT/DM  is  equal  to  19'5/41'2;  in  this  way 
the  point  T  has  been  found.  It  is  then  only  necessary  to  find  the 
centre,  F,  of  the  circle  passing  through  D,  T  and  0  to  complete 
the  construction  for  the  three  circles. 

It  will  be  remembered  that  the  line  on  which  the  slip  was 
measured  was  drawn  through  M,  the  reason  for  choosing  M  'in 
preference  to  any  other  point  will  now  be  evident.  For  at  M  the 
rotor  is  at  rest  and  hence  the  slip  is  100°/0,  hence  the  line  MS  can 
be  divided  as  shown  into  a  number  of  equal  parts  so  that  on  it  the 
slip  as  a  percentage  of  the  speed  of  the  field  can  be  read  off  for 
any  position  of  the  current  vectors.  Thus  the  distance  Ss  is  not 
merely  proportional  to  the  slip,  but  measures  directly  the  per- 
centage slip  occurring  for  the  stator  current  OP. 

Determination  of  scales.  We  will  now  see  how  to  deduce 
the  complete  performance  of  the  motor  from  light  load  up  to  the 
point  of  maximum  power  factor.  The  scale  of  Fig.  165  is  too  small 
for  the  different  quantities  to  be  measured  on  it,  and  in  Fig.  166  is 
given  the  part  of  the  circles  for  the  desired  range  of  load  with  a 


Fig.  166. 


ram  all  the 
torque  can 


scale  of  one  centimetre  per  ampere.     From  this  di 

necessary  quantities  except  the  slip  and  the  full  lo; 

be    found.      These    quantities    were    measured    in    the    manner 

described  on  the  complete  circles,  but  these  manifestly  cannot 

be  reproduced. 

The  first  point  to  be  considered  is  the  scales  on  which  the 
different  quantities,  input,  output  and  torque,  must  be  measured. 
A  scale  of  current  is  shown  along  which  currents  can  be  measured. 


THE   HEYLAND   CIRCLES  201 

The  end  point  D  of  the  diagram  is  in  the  direction  shown  by  the 
arrow. 

The  input.  This  is  evidently  given  by  the  projection  of  the 
stator  current  line  OP  on  the  direction  of  applied  pressure  OF, 
as  shown  on  p.  197.  But  the  value  of  the  pressure  per  phase  is 

120 

—  —  volts  and  hence  the  total  power  supplied  by  the  three  stator 

windings  is  120  v/3  Pp  watts  or  0'208  Pp  kilowatts. 

The  output.  On  the  same  page  we  saw  that  the  nett  output 
was  equal  to  the  nett  projection  of  the  rotor  current  multiplied 
into  the  pressure  represented  by  the  line  CD.  But  if  a  is  the 
constant  of  the  motor  referred  to  on  p.  193  we  evidently  have  that 

CP 

the  total  rotor  current  is  -   -  and  hence  the  nett  inphase  rotor 

current  is  -—  .     But  we  there  saw  that  the  relation  between  the 
a 

lines  CD  and  DO  was  given  by  DC=z.DO  and  the  latter  represents 
the  applied  pressure  of  120/V3  volts.  It  follows  that  the  output 
in  watts  of  the  rotor,  taking  into  account  the  three  phases  of  the 
machine,  will  be  120  \/3  .  Rr,  or  in  horse-powder  is  given  by 


The  torque.  The  rate  of  transmission  of  energy  from  the 
stator  to  the  rotor  will  be  given,  by  what  was  seen  on  p.  197,  by 
the  product  Qq  into  CD.  But,  as  in  the  last  case,  this  inphase 
current,  being  in  the  rotor  windings  and  reacting  there  with  the 
flux,  must  be  interpreted  on  the  rotor  current  scale.  Hence  the 
value  of  this  power  is  given  by  the  product  of  the  nett  inphase 
current,  given  by  the  length  of  Qq,  and  the  pressure  given  by  OD. 
Hence  the  value  of  the  power  per  phase  transferred  across  the 
air  gap  will  be  OC  .Qq  or  120\/3.Qg  watts,  taking  as  before  all 
three  phases  into  account.  In  order  to  determine  the  torque  it  is 
necessary  to  express  the  pressure  in  terms  of  some  definite  flux 
and  the  angular  velocity  of  the  field.  The  latter  is  running  round 
at  the  synchronous  speed  of  43/2  revolutions  per  second,  since  the 
field  of  the  stator  is  four  pole.  Hence  the  angular  velocity  is  43-Tr  or 
135  radians  per  second.  It  follows  that  the  torque  in  units  such  that 
work  is  in  joules  and  angular  velocity  in  radians  per  second  will  be 

given  by  -  -~—  Qq.    Further,  to  reduce  this  torque  to  foot-pound 

loO 

units  we  must  multiply  by  f  |g,  hence  in  this  case  the  torque  can 

120\/3x550 
be  found  from  the  length  of  Qq  by  multiplying  it  by  —          , 

or,  torque  in  foot-pound  units  =  1*14  Qq. 


202 


ALTERNATING  CURRENTS 


The  efficien 

the  efficiency  is  r 

The  power 

variable  is  the  t 
in  each  case  of  tt 

2-5     10    WO 

»J* 

2  '  £8  $80 

£  ^7';> 
g  §  g 

IV»- 

Ql-5^4-^4C 
*$•     <b    ^ 

0-j    i  .$t>26> 

cy.     Evidently  from  the  above  it  will  follow  that 
nerely  the  ratio  of  Rr  to  Pp. 

factor.    If  the  quantity  taken  as  the  independent 
otal  current  passing,  the  power  factor  is  the  ratio 
ie  line  Pp  to  this  current. 

/ 

/ 

F^ 

au- 

-7 

/ 

A 

rf^x' 

r 

< 

Sfc« 

ncy 

7 

/ 

/f 

/* 

^ 

r 

// 

x 

/? 

/ 

-, 

^V 

? 

/ 

X 

1 

x 

/ 

/ 

Y 

z 

O  O-5  1-C  1-5  2-0  2-5 

Output  in  Horse-power. 

Fig.  167. 

Hence  if  the  necessary  lines  are  drawn  for  any  desired  currents 
say  from  5  to  12  amperes  as  in  this  case,  and  if  a  table  be  drawn 
up  of  the  three  lines  Pp,  Qq  and  Rr,  the  whole  performance  of 
the  motor  can  be  at  once  determined.  This  has  been  done  as 
shown  in  the  table  given  below,  the  results  being  plotted  in  the 
curves  shown  in  Fig.  167. 


Current      Pp 

4-1 

5 

6 

7 

8 

9 
10 
11 
12 


2-83 
4-22 
5-41 
6-54 
7-57 
8-58 
9-61 
10-56 


T79 
3-14 
425 
5-30 
6-25 
7-15 
8-09 
8-91 


Er 

1-72 
3-00 
4-04 
4-95 
5-82 
6-60 
7-40 
8-10 


Input 

Output 

in  K.W. 

in  H.P. 

0-24 



0-59 

0-48 

0-88 

0-84 

1-12 

1-13 

1-36 

1-38 

1-57 

1-62 

1-77 

1-84 

1-98 

2-03 

2-20 

2-26 

61 
71 

74 
75 
76 
76 

77 


Torque, 

Cos  X    foot-pound 
°/o  units 

25 
57 
70 
77 
82 
84 
86 
87 
88 


2-02 
3-58 
4-85 
6-05 
7-15 
8-15 
9-25 
10-02 


CHAPTER  XVI. 

EQUATIONS  FOR  INDUCTION  MOTOR. 

Ix  Chap.  XV  we  saw  that  the  determination  of  the  no-load 
power  and  current  at  normal  pressure,  and  the  determination  of 
the  stand-still  load  and  current  at  a  reduced  pressure,  enabled  us 
to  derive  a  graphical  construction  by  which  the  whole  performance 
of  a  transformer  could  be  predicted.  Again,  in  Chap.  V  we  derived 
an  algebraic  method  of  treating  the  ordinary  transformer  from 
similar  results.  We  will  now  develop  a  similar  algebraic  method 
by  which  the  performance  of  an  induction  motor  can  be  found  from 
observations  of  this  type. 

In  what  follows  the  current  and  volts  will  be  specified  by  their 
virtual  values  to  avoid  the  introduction  of  factors,  and  they  will  be 
assumed  to  be  sinusoidal  as  before.  In  the  two  tests  considered, 
the  no-load  and  the  stand-still,  the  power  generally  measured  is 
that  given  to  the  whole  stator,  while  the  pressure  is  measured 
between  two  adjacent  mains,  the  current  being  that  in  one  main. 
Thus  if  the  motor  be  a  two-phase  one,  the  power  per  phase  will  be 
one-half  of  the  measured  power,  while  the  current  delivered  by 
each  phase  and  the  pressure  between  the  phases  can  be  directly 
measured.  In  the  case  of  a  three-phase  motor,  the  power  per 
phase  will  be  one-third  of  the  total  power ;  if  star-connected  the 
current  in  the  mains  will  be  equal  to  the  current  in  the  winding 
of  the  stator  while  the  pressure  between  two  mains  will  be  \/3 
times  the  pressure  across  one  winding.  In  the  mesh  case,  the 
pressure  between  the  mains  will  be  the  same  as  that  across 
a  winding  of  the  stator,  while  the  current  in  the  mains  will  be 
>v/3  times  the  current  in  the  winding.  In  what  follows  the 
symbols  for  the  respective  pressures,  currents  and  power  will  refer 
to  a  single  winding  of  the  stator,  and  must  therefore  be  deduced 
from  the  observed  readings  by  the  above  factors.  The  letter  7 
will  be  used  to  denote  the  number  of  stator  circuits,  thus  7=2 
for  a  two-phase  stator,  and  7  =  8  for  a  three-phase  one. 

No-load  or  light-load  test.  Let  an  induction  motor  be  run 
at  no  load,  that  is,  with  no  mechanical  load  on  the  rotor,  and  let  dif- 
ferent pressures  be  applied  to  the  stator  at  the  proper  periodicity, 


204 


ALTERNATING    CURRENTS 


starting  with  the  maximum  pressure  for  which  it  is  designed,  and 
let  the  current  and  power  per  circuit  be  plotted  against  the  volts 
on  the  terminals  of  that  circuit.  Two  curves  will  be  obtained  of 
the  form  shown  in  Fig.  168.  0V  is  the  full  pressure,  and  VW  and 
VC  are  the  corresponding  power  and  current.  It  will  be  found 
that  the  motor  will  stop  rotating  at  a  definite  value  of  the  pressure. 
This  occurs  because  the  torque  produced  is  at  that  point  only  just 
sufficient  to  make  up  for  the  internal  losses,  at  the  constant 
revolutions  of  the  rotor  consequent  on  the  constant  applied 
periodicity;  it  will  be  recollected  that  on  p.  184  it  was  shown 


0  Volts  V 

Fig.  168. 

that  the  maximum  torque  is  roughly  proportional  to  the  square 
of  the  applied  pressure.  The  ordinates  of  the  curve  WW  give  the 
power  lost  corresponding  to  the  given  applied  pressures.  This 
consists  of  the  following  parts :  the  frictional  loss  which  will  be 
constant  since  the  speed  is  constant,  the  core  loss  in  the  machine, 
and  the  ohmic  losses  in  the  stator  and  rotor.  Since  the  torque 
produced  by  the  rotor  is  small,  the  current  flowing  in  it  is  also 
small,  and  hence  the  ohmic  loss  in  the  rotor  is  negligibly  small, 
thus  the  total  losses  are  practically  the  core  loss,  friction  loss,  and 
the  stator  ohmic  loss.  The  resistance  (per  phase)  of  the  stator, 
which  we  will  denote  by  H0,  can  readily  be  measured  in  any  of  the 
ordinary  ways  at  its  normal  working  temperature,  and  hence  for 
any  applied  pressure  in  the  curve  the  corresponding  stator  current 
can  be  read  off  from  the  curve  CC  and  the  corresponding  ohmic 
loss  deducted  from  the  observed  total  loss  of  power,  we  thus 
readily  derive  the  curve  LL  giving  the  loss  in  friction  and  that  in 
the  core  for  the  different  pressures  applied.  Let  LL  be  produced 
to  cut  the  axis  in  P  which  can  always  be  done  with  fair  accuracy, 
and  it  will  be  seen  that  OP  represents  the  constant  frictional  loss. 
Hence  if  the  parallel  line  PF  be  drawn,  cutting  the  vertical  drawn 
through  the  full  working  pressure,  the  line  VF  will  give  the 
frictional  loss  at  full  load,  and  LF  will  give  the  corresponding  core 


EQUATIONS   FOR   INDUCTION   MOTOR  205 

loss,  the  sum,  or  VL,  is  the  total  loss  incident  to  the  rotation  of 
the  rotor,  apart  from  any  ohmic  loss,  and  we  will  denote  it  by  W0. 

Now  let  ^0  be  the  no-load  current  at  full  pressure  as  read 
from  the  upper  curve,  that  pressure  being  throughout  denoted 
by  SQ  and  being  constant  in  value  and  periodicity,  the  power 
component  of  the  no-load  current  will  be  given  by  TFo/do  and 
will  be  denoted  by  ^p.  Consequently  the  wattless,  or  quadrature, 
component  will  be  given  by  (^02  —  ^2)*  and  will  be  denoted  by  ^q. 

The  no-load  test,  therefore,  enables  us  to  find  the  frictional 
losses,  the  core  losses,  and  the  two  constant  components  of  the 
current  required  to  maintain  the  field  and  make  up  for  the 
rotational  losses,  that  is  ^p  and  ^q.  The  former  is  always  to  be 
taken  in  phase  with  the  terminal  pressure,  the  latter  at  right 
angles  thereto. 

Effect  of  want  of  phase  balance.  In  cases  where  the  im- 
pressed pressure  vectors  are  not  at  the  proper  phase  angle,  i.e.  90° 
for  2-phase  and  1 20  for  3-phase,  the  resulting  want  of  balance  may 
cause  very  serious  differences  in  the  power  taken  by  the  phases. 
For  example  the  numbers  given  below  refer  to  a  test  of  a  two- 
phase  motor  at  no  load  in  which  the  alternator  had  the  two 
E.M.F.S  slightly  out  of  quadrature  owing  to  its  possessing  a  closed 
winding  with  four  tappings  which  were  not  such  as  to  include 
exactly  the  same  number  of  conductors  in  each  quadrant.  It  will 
be  seen  that  the  power  taken  by  the  two  phases  is  very  unequal 
and  in  fact  that  taken  by  one  of  the  phases  actually  diminishes 
with  increase  of  pressure  instead  of  increasing  in  the  normal 
manner.  The  reason  for  this  effect  is  that  the  rotor  currents  tend 
to  establish  a  neutral  point  by  reacting  on  the  stator,  and  this  point 
is  not  the  same  as  that  impressed  on  the  stator  windings.  Or  we 
may  say,  that  taking  one  phase  of  the  stator  for  reference,  the 
other  phase's  E.M.F.  can  be  looked  upon  as  possessing  a  component 
exactly  in  quadrature  with  the  first  one,  and  an  outstanding 
component.  This  latter  will  send  current  round  the  circuit  which 
will  be  superposed  on  the  true  two-phase  ones,  and  lead  to  a  want 
of  balance.  The  phase  relations  are  such  that  little  difference 
exists  between  the  currents  in  the  two  phases.  If  the  exact 
quadrature  of  the  impressed  pressures  cannot  be  secured,  the  test 
must  be  carried  out  with  the  power  measured  in  both  of  the 
phases. 

Impressed  Power  in  watts,       Power  in  watts, 

pressure  Phase  A  Phase  B 

407  94  586 

350  100  500 

310  106  390 

260  117  330 

200  130  390 


206 


ALTERNATING   CURRENTS 


Locked   characteristic    or   stand-still   curve.      Let    the 

rotor  be  blocked  so  as  to  be  incapable  of  rotation,  and  different 
pressures  be  applied  to  the  stator,  in  each  case  reading  the 
corresponding  current  and  power  per  phase.  Curves  will  be 
obtained  like  those  in  Fig.  169,  the  current  one  being  practically 
a  straight  line.  In  this  test,  since  the  rotor  is  prevented  from 
moving,  the  friction  losses  are  zero,  but  there  will  be  certain 
core  losses  in  the  machine.  Up  to  values  of  the  current  not 


Volts 

Fig.  169. 

exceeding  very  much  the  full-load  current,  the  terminal  pressure 
on  the  stator  will  be  only  a  fraction  of  the  normal  working  pres- 
sure, and  hence  such  core  losses  will  be  due  to  only  a  fraction 
of  the  full  induction  cycle  in  the  iron.  Thus  in  comparison  with 
the  ohmic  losses  they  will  be  small.  Even  if  pressures  of  greater 
amount  are  used,  the  losses  due  to  resistance  will  evidently  be 
many  times  the  corresponding  core  losses  for  the  pressures  used, 
and  hence  the  losses  in  this  test  can  be  very  approximately  taken 
as  being  solely  ohmic  in  character.  Further  it  also  follows  that 
the  currents  flowing  in  the  stator  to  produce  the  flux  will  in  every 
case  be  very  small  in  comparison  with  the  induced  rotor  current, 
and  hence  the  stator  current  belt  can  be  taken  as  being  practically 
the  same  as  that  in  the  rotor.  In  fact  this  test  is  practically 
equivalent  to  the  short  circuit  test  of  a  transformer  given  on 
p.  68,  and  can  be  considered  in  the  same  way.  Thus  we  may 
replace  the  actual  resistances  of  stator  and  rotor  by  an  equivalent 
stator  resistance,  and  the  E.M.F.S  due  to  the  leakage  fields  of  stator 
and  rotor  can  be  replaced  by  reactances  of  the  amount  necessary 
to  produce  the  same  quadrature  E.M.F.S  as  the  actual  leakage 
fields.  The  resistance  of  the  stator  can,  as  has  been  said, 
be  measured  directly  and  was  denoted  by  R0 :  let  the  actual  re- 
sistance of  the  rotor  be  equivalent  under  the  stand-still  conditions 
to  a  resistance  of  the  amount  ^  in  the  stator,  then  the  whole 
apparatus  acts  as  if  it  had  the  ohmic  resistance  (E0  +  J^).  If  the 
rotor  is  a  wound  one,  the  true  resistance  of  its  coils  can  be  directly 


EQUATIONS    FOR   INDUCTION    MOTOR  207 

measured  in  the  ordinary  way,  and  this  resistance  can  be  reduced 
to  the  equivalent  stator  resistance  in  the  method  described  in 
connection  with  the  transformer,  when  the  form  and  turns  of  the 
two  windings  on  the  stator  and  rotor  are  known.  In  the  case 
of  a  short-circuited  rotor  this  cannot  well  be  done,  but  we  shall  see 
that  the  stand  -still  test  will  enable  us  to  find  the  value  of  the 
equivalent  resistance  of  the  rotor  Rlf 

It  must  be  noted  that  since  in  this  case  the  rotor  and  stator 
currents  are  the  same  (since  the  magnetising  current  is  negligible) 
with  a  stator  current  ^  the  ohmic  drop  will  be  r^>g(R0  +  Rl)  at 
stand-still. 

Now  consider  the  reactance  pressures,  and  let  SQ  denote  the 
reactance  of  the  stator  at  the  normal  periods,  with  the  current  ^g, 
the  corresponding  quadrature  pressure  will  be  ^$0.  Similarly  let 
$!  be  the  rotor's  reactance  at  full  periods,  as  in  the  stand-still 
test,  then  its  equivalent  quadrature  pressure  will  be  (^>SS1.  Thus 
the  total  quadrature  pressure  will  be  ^f«  OS>0  +  $1)-  Hence  if  S8 
and  ^s  denote  the  observed  stand-  still  pressure  and  current, 
is  the  impedance  of  the  circuit  or  /.  We  then  evidently  have 


The  fact  that  the  current-pressure  curve  is  a  straight  line 
shows  that  these  quantities  are  approximately  constants  for  the 
machine. 

But  the  power  test  enables  us  to  separate  the  resistance  and 
reactance,  for  if  W8  is  the  stand-still  power  with  the  current  ^ 
we  must  have  Wg  =  <$?  (R0  +  J^),  which  gives  the  value  of 
(RQ  -f-  RI)  and  thus  enables  us  to  find  the  value  of  ($0  +  8^. 
Also,  since  the  value  of  R0  has  been  obtained,  we  can  find  the 
equivalent  rotor  resistance  or  Rl.  Thus  from  the  stand-still  test 
we  derive  the  equivalent  rotor  resistance  and  the  sum  of  the 
reactances  under  the  given  condition.  The  latter  cannot  be 
separated  into  the  parts  ^  and  S0  in  the  test,  but  in  all  the 
practical  applications  of  the  results  this  separation  is  unnecessary. 

Equivalent  resistance  and  reactance  of  stator  when 
rotation  occurs.  Fractional  slip.  We  must  now  consider 
what  occurs  on  rotation  being  permitted,  and  for  this  purpose  we 
will  alter  the  manner  in  which  the  slip  is  specified.  Let  us  write 
a-  =  %7rn  .  2  =  2  .  fl,  then  2  is  the  fractional  slip,  or  when  ex- 
pressed as  a  percentage,  the  percentage  slip.  At  synchronism 
it  is  zero,  and  at  stand-still  it  is  unity.  In  what  follows  to  save 
trouble  it  will  be  called  the  slip  only.  Then  the  angular  velocity 
of  the  rotor,  of  a  two-pole  motor,  will  be  given  by 

o>  =  fl  —  cr   or   a)  =  fl  (1  —  2). 

As  in  the  transformer,  let  us  suppose  only  the  load  currents, 
*$,  to  be  flowing,  consisting  of  two  equal  belts  in  rotor  and 


208  ALTERNATING   CURRENTS 

stator,  and  find  the  relation  between  the  current  and  pressure 
in  the  stator  with  a  slip  2  existing;  the  stator  must  for  this 
purpose  be  looked  upon  as  a  mere  choking  coil,  in  the  same  way 
as  the  transformer  was  treated  in  Chapter  V. 

As  far  as  the  two  leakage  fields  are  concerned  the  rate  of 
cutting  is  the  same  as  in  the  stand  -still  case,  hence  the  quadrature 
pressure  will  be  still  given  by  *$  (S0  -f  $j)  where  *$  is  the  current 
flowing,  which  is  the  same  as  in  the  rotor.  As  regards  the 
equivalent  resistance  the  case  is  different.  The  true  stator 
resistance,  RQ,  is  unaltered,  but  we  shall  see  that  resistance  in  the 
stator  that  is  now  equivalent  to  the  rotor's  resistance  will  be 
^  divided  by  2.  Since  all  the  leakage  pressures  have  been 
transferred  to  the  stator,  the  sole  pressure  that  is  left  for 
consideration  in  the  rotor  is  that  for  its  ohmic  drop  which  has  the 
value  yj>Ri.  Now  when  rotation  takes  place  the  common  flux  will 
generate  pressures  in  both  circuits,  and  the  relative  values  of  the 
pressures  will  be  as  the  corresponding  periodicities.  Let  x  denote 
any  pressure  so  produced  in  the  rotor,  its  periodicity  at  the  slip  2 
will  be  'S.n  where  n  is  the  impressed  periodicity  ;  hence  the 

np 

equivalent   pressure   in   the   stator   will    be  ^  since   there   it    is 

generated  at  full  periodicity,  it  follows  that  if  the  pressure  in  the 
rotor  due  to  drop  is  ^.R^  the  equivalent  one  in  the  stator  is 
and  thus  the  equivalent  total  stator  ohmic  drop  at  slip  X  is 


f  -p    ,  RI\ 

(    °  +  Ig  J 


Hence  all  the  pressures  that  can  exist  in  the  stator  are  the 
following.     One  of  the  value  ^  f  R^  4-  ^-M  and  one  of  the  value 

9!?  (Si  +  $o)  where  in  each  case  *$  is  the  current  flowing.  These 
two  are  in  quadrature  and  thus  the  total  pressure  in  the  stator 
must  be  the  square  root  of  the  sum  of  the  squares  of  these  two 
components.  But  this  must  also  be  the  value  of  the  constant 
applied  stator  pressure  per  phase,  and  hence  we  finally  get 


This  result  should  be  compared  with  that  arrived  at  in  the  case 
of  the  transformer.  It  will  be  seen  to  agree  with  it  in  every 
detail.  Possibly  the  absence  of  any  reference  to  a  back  or  induced 
E.M.F.  in  the  present  case  may  prove  a  point  of  difficulty,  but 
it  must  be  recollected  that,  regarded  as  a  purely  electrical  device, 
there  is  no  external  secondary  pressure  to  consider,  the  equivalent 
to  this  is  the  mechanical  output. 

Resistance  and  reactance  of  Rotor  when  rotation 
occurs.  The  next  point  is  to  see  how  the  current  belt  in  the  rotor 


EQUATIONS   FOR   INDUCTION   MOTOR  209 

(which  is  equal  to  that  in  the  stator)  is  related  to  any  induced 
pressure  existing  in  the  rotor.  Let  £r  be  any  such  E.M.F. ;  we  must 
first  find  an  expression  for  the  impedance  of  the  circuit  of  the 
equivalent  rotor;  when  it  is  running  the  resistance  will  be  still  Rl} 
but  the  reactance  will  be  reduced.  For  the  quantity  $,  denotes 
the  pressure  produced  when  the  periodicity  is  n  and  the  current  is 
unity,  hence  when  the  current  is  still  unity  but  the  periodicity 
is  less,  the  pressure  produced  will  be  less  in  proportion.  Let  the 
slip  be  2,  then,  instead  of  the  periodicity  of  the  currents  in  the 
rotor  being  n  it  is  2n,  and  thus  if  the  rotor  be  carrying  any 
current  <$  at  that  slip,  the  reactance  pressure  will  be  ^.2.$!. 
But  with  the  same  current  the  ohmic  drop  is  ^ .  Rl  and  hence  the 
total  drop  is  <@(R?  +  ¥.Srf.  It  follows  that  the  impedance 
of  the  rotor's  circuits  at  the  slip  2  will  be  given  by  (Rf  +  22 .  $x2)^ 
Hence  when  any  pressure  £r  exists  in  the  rotor,  the  current 
(in  either  rotor  or  stator)  will  be  given  by 


Relation  between  impressed  stator  pressure  and  in- 
duced rotor  pressure.  We  can  now  find  the  necessary  relation 
between  the  impressed  stator  pressure  and  the  corresponding 
value  of  Sr  for  any  assigned  value  of  2,  for  from  the  rotor  side 
the  current  is  given  by 


+  22  .  S?  ' 
while  from  the  stator  side  it  is  given  by 


It  follows  that  Sr  is  given  in  terms  of  ^0  by  the  expression 


v  (§ 


...         . 

or  &r=    ,  ...............  (!)• 

+  2 


Torque  and  output.  The  next  point  is  to  deduce  expressions 
for  the  torque  and  output  of  the  motor. 

Suppose  that  it  is  of  the  ordinary  two-pole  type,  let  P  be  the 
torque  it  produces  :  then  the  waste  of  energy  in  the  rotor's  wires 
will  be  Per,  where  or  has  the  former  meaning  of  the  true  slip  as  an 


14 


210  ALTERNATING   CURRENTS 

angular  velocity.  For  if  there  were  no  such  ohmic  losses,  the 
rotor  currents  would  require  no  pressure  to  produce  them,  and  thus 
no  slip  would  be  required,  and  the  angular  velocity  of  the  rotor 
would  be  n  (the  same  as  the  field),  while  the  rate  of  working 
would  be  PO.  But  the  actual  rate  of  doing  work  is  Po>,  and  hence 
the  energy  lost,  incident  to  the  passage  of  the  rotor  current  on  its 
wires,  is  P  (£1  —  a>)  or  Pa.  This  must  be  equal  to  the  ohmic  loss 
in  the  rotor.  We  have  seen  that  the  current  that  circulates 
in  any  one  of  the  rotor's  circuits  is 


and  hence  with  7  circuits  on  the  stator  the  equivalent  rotor  ohmic 
loss  will  be  given  by  the  square  of  the  current  in  each  circuit 
multiplied  by  the  corresponding  resistance  fr  and  the  number  of 
stator  circuits.  Also  with  our  new  notation  for  the  slip,  we  have 
2  .  27r?i  =  cr.  Hence 

p  a 

P. 


substituting  for  the  value  of  Sr  in  terms  of  SQ  from  (1)  we  get 

P  =  _  ffl.S.J^.y  . 

-  -  >     ' 


The  units  in  which  this  torque  is  measured  will  be  those 
corresponding  to  joules  for  work  and  radians  for  angular 
velocity. 

We  can  now  deduce  the  power  that  the  rotor  is  delivering. 
For  we  have  its  angular  velocity,  o>,  given  by 

co  =  fl  —  cr,    or    o>  =  %7rn  (1  —  S). 

Hence  the  power  in  watts  will  be  the  product  of  the  torque  and 
this  angular  velocity,  or  will  be  given  by 

ffl.fr.  7.  2(1-2) 

»       «  '" 


It  will  be  seen  from  Chap.  XVII  that  if  the  motor  has  H  pairs  of 
poles,  since  the  power  will  be  the  same,  but  the  angular  velocity  is 

yjth  of  its  value  for  two  poles,  the  torque  produced  will  be  H  times 

the  expression  in  equation  (2). 

The  following  deductions  can  be  made.     In  expression  (2)  if  we 

7  r> 

equate  -j=-  to  zero,  it  will  give  the  slip  at  which  maximum  torque 
occurs  ;  this  will  be  found  to  lead  to 

T> 

P  ~ 


EQUATIONS   FOR   INDUCTION   MOTOR  211 

The  corresponding  torque  will  be  given  by  substituting  this 
value  of  2p  in  equation  (2).  The  starting  torque  will  evidently  be 
given  by  equation  (2)  if  S  is  taken  equal  to  unity.  This 
leads  to 

A'.fr.y 
27m  2" 


But  the  quantity  in  the  bracket  is  evidently  the  (impedance)2  at 
stand-still,  and  is  hence  the  value  of  that  quantity  as  found  in  the 
stand-still  test.  But  ^02  divided  by  the  bracket  must  therefore  be 
the  square  of  the  stand-still  current  at  full  impressed  pressure. 
But  this  multiplied  by  7  .  1^  is  the  energy  lost  in  the  whole  rotor, 
and  hence  we  see  that  the  starting  torque  is  given  by  the  whole 
loss  in  the  rotor  divided  by  the  angular  velocity  of  the  field. 
This,  in  fact,  follows  from  the  consideration  on  p.  209. 

The   slip   for   maximum   work   can  also  readily  be  found  by 

equating  -^-  to  zero  :  it  will  be  found  to  lead  to  the  expression 


and  if  this  expression  for  the  slip  be  substituted  in  equation  (3) 
we  find 

/p  g 

maximum  W=  -          -  ,7     "  .  .  .(6). 

R.)  +  VCR,  +  R,Y  +  (S, 


We  thus  see  that  a  knowledge  of  the  value  of  the  quantities 
R0,  jRj  and  ($!  +  S0),  as  given  by  the  stand-still  test  and  the  direct 
measurement  of  jK0,  will  enable  us  to  find  the  torque  at  any 
assigned  value  of  2  from  equation  (2),  and  the  power  from 
equation  (3)  ;  they  will  also  enable  us  to  predict  the  maximum 
torque  by  means  of  equations  (4)  and  (2),  the  starting  torque  from 
equation  (5),  and  the  maximum  power  delivered  by  the  rotor  from 
equation  (6). 

Expression  for  stator  current.  We  will  now  proceed  to 
see  how  an  approximate  expression  can  be  found  for  the  stator 
current  at  any  slip  in  the  region  where  this  slip  is  small,  say  not 
greater  than  6  °/0  ,  that  is,  in  the  region  which  is  of  importance  in 
the  operation  of  the  motor. 

The  current  flowing  in  the  stator  consists  of  two  parts,  the 
magnetising  current  and  the  load  current.  By  the  no-load  test  we 
saw  that  we  could  determine  the  components  of  the  former  current, 
namely,  ffip  in  phase  with  the  applied  pressure,  and  ^q  in  quadra- 
ture therewith. 

The  load  current  which,  as  we  have  said,  is  the  same  for  rotor 
and  stator,  can  be  similarly  divided  into  two  components,  the  power 
component  in  phase  with  the  pressure,  and  the  wattless  component 

14—2 


212  ALTERNATING   CURRENTS 

in  quadrature  therewith.     If  &r  denote  as  before  the  rotor  E.M.F. 
the  rotor  current  will  be 


This   will   lag   on   the   pressure  by  the  angle  whose  tangent  is 

2jS 

^^.     Hence  the   power  component   will   be   the   value   of    the 

jftj 

current  multiplied  by  the  cosine  of  this  angle,  or  by 


while  the  wattless  component  will  be  the  current  multiplied  by  the 
sine  of  the  angle,  or  by 


Consider  the  power  component,  its  value  will  be 


Now  substitute  for  £r  its  value  in  terms  of  SQ  given  by  equation  (1) 
and  we  get  that  the  power  component  is 


VlC*  +  2  • 

This  holds  for  any  value  of  2,  but  for  the  running  part  of  the 
mechanical  characteristic  2  is,  as  we  have  said,  small ;  hence  if  we 
neglect  its  square  in  comparison  with  the  other  quantities,  we  have 
that  the  power  component  is  given  by 


The  total  power  component  is  the  sum  of  this  and  ^p  .    Hence  the 
total  power  component  of  the  stator  current  is 


>    ,  - 

2.  2. 


Now  consider  the  wattless  component  of  the  working  current. 
By  similar  substitutions  we  shall  find  it  is  given  by 


2  . 

or,  if  we  neglect  in  the  same  way  terms  affected  by  22,  it  is  to  be 
left  out  of  account.  Hence  the  sole  wattless  component  of  the 
stator  current  is  the  flux  component  of  the  no-load  or  magnetising 
current,  that  is  ffi. 


EQUATIONS   FOR   INDUCTION  MOTOR  213 

But  the  stator  current  will  be  the  root  of  the  sum  of  the 
squares  of  these  two  components,  hence  for  any  assigned  value  of 
the  quantity  2  it  is  given  by 


......  (8)- 

The  value  of  the  starting  current,  when  2  =  1,  can  be  readily 
found.  It  is  merely  the  result  of  dividing  the  pressure  of  supply 
by  the  impedance  found  in  the  stand-still  test,  or  is 


.(9). 


Prediction  of  performance.  The  complete  performance  of 
the  motor  can  now  be  predicted  from  a  measurement  of  the  stator 
resistance  per  phase,  the  deduction  of  the  corresponding  rotor 
resistance  and  the  combined  stand-still  reactance  from  the  stand- 
still curve,  together  with  the  determination  of  the  load  component 
and  flux  component  of  the  magnetising  current  from  the  no-load 
test. 

The  starting  torque,  maximum  torque,  starting  current  and 
maximum  power  can  be  immediately  calculated  from  equations  (5), 
(4  with  2),  (9)  and  (6)  respectively.  The  current  taken  can  also  be 
found  in  relation  to  the  slip  up  to  a  value  of  the  sums  corre- 
sponding to  about  6  %  of  full  frequency.  Assume  a  set  of  values 
for  2  and  for  each  calculate  the  current  from  equation  (8)  and  the 
corresponding  torque  and  power  from  equations  (2)  and  (3). 
Curves  exhibiting  the  relation  between  the  torque  as  abscissa  and 
the  current  per  phase  and  the  power  can  then  be  drawn.  The 
slip  can  evidently  be  also  drawn  in  at  each  value  of  the  torque, 
since  it  is  the  assumed  quantity.  The  losses  can  be  found  as 
follows:  for  any  one  stator  current  the  corresponding  loss  per 
phase  will  be  the  product  of  the  square  of  that  current  into  the 
resistance  of  the  stator's  winding ;  the  loss  in  the  rotor  will  be  the 
product  of  the  square  of  the  load  current  into  the  equivalent  rotor 
resistance  ;  the  total  core  and  rotational  loss  has  been  derived  from 
the  no-load  test,  and  hence  the  total  loss  at  any  assumed  value  of 
the  slip  can  be  derived.  This  added  to  the  calculated  output  for  the 
assumed  slip  will  give  the  corresponding  input,  and  the  ratio  of  the 
former  to  the  latter  is  the  true  efficiency.  Instead  of  plotting  the 
current  per  phase  it  is  often  customary  to  plot  the  total  apparent 
power  taken  by  the  motor,  that  is  to  say  the  product  of  the  applied 
pressure  per  phase  into  the  sum  of  the  currents  in  the  phases; 
since  the  motor  has  been  assumed  to  be  a  balanced  load  on  the 
mains,  the  power  factor  will  evidently  be  found  by  dividing 
the  real  power  by  this  apparent  power  at  each  value  of  the 
torque.  Thus  a  set  of  curves  can  be  drawn  up  which  will  fully 
exhibit  the  performance  of  the  motor. 


214 


ALTERNATING   CURRENTS 


In  Figs.  170  and  171  are  given  the  no-load  and  stand-still  test 
curves  of  a  two-phase  motor  designed  for  a  terminal  pressure  of 
400  volts,  a  periodicity  of  60,  and  an  output  of  5  h.-p.  at  900 
revolutions  per  minute.  The  number  of  poles  has  to  be  8  to  fulfil 


275 


225 

200 
175 

L 


100 
75 


25 


6-0 
5-5 

5-0 


C  O          50     100       ISO     200      250     300     350     400     450     500 

Fig.  170. 

the  conditions.  The  resistance  of  the  stator  was  found  to  be  3'79 
ohms.  The  curves  are  drawn,  as  is  customary,  for  the  total  current 
and  total  watts  taken  by  the  motor,  so  that  they  should  be  avail- 
able for  comparison  of  motors  whether  of  two  or  three  phases. 
On  referring  to  the  stand-still  test  it  will  be  seen  that  the 
current  per  phase  at  a  pressure  of  400  volts  is  22 -1  amperes  and 
the  power  is  4600  watts.  We  deduce  that  the  equivalent  total 
stator  resistance  is  4600-=- (22*1  )2  ohms  or  9'41  ohms,  hence  the 
equivalent  rotor  resistance  is  the  difference  between  this  and  the 
stator  resistance,  or  is  5*62  ohms,  we  thus  get 

#0  =  379,     #1  =  5-62. 

The  total  stand-still  impedance  is  400-f-22'l  or  18'1  and  hence  the 
stand-still  reactance  is 


or  \/237-5. 


Hence  we  have 


From  these  data  the  power,  etc.  can  be  calculated  as  above  described. 
For  example  take  the  case  where  the  slip  is  9t2°/0,  the  value 


EQUATIONS   FOR   INDUCTION   MOTOR 


215 


corresponding    to    maximum    output.      The   expression   for   the 
output  is  given  in  equation  (3);   we  have 

<£o=400,     ^  =  5-62,     7=2, 

2  (1  -  2)  =  0-092  x  -908  =  0'0835, 
(R,  +  S^o)2  =  (5-62  +  0-35)2  =  35'6, 
22  (S,  +  S0)2  =  0-0085  x  237-5  =  2'0. 
Hence  on  substituting  in  the  equation 


we  get 


W 


w= 


watts 


160  x  5-62  x  2  x  0'0835 


,  ., 

kilowatts, 


or 


35-6  +  2 
F=  4-0  kilowatts  =  5'3  h.-p. 

The  speed  can  be  immediately  deduced  from  the  assumed  slip; 
since  the  synchronous  speed  is  900  R.P.M.  it  will  be  0'908  of  this 
or  817  R.P.M. 

From  the  open  circuit  curve  we  deduce  the  power  component 
*$p  and  the  powerless  component  ^q.     Under  these  circumstances 


j 

35 
30 

$20 
IS 
W 

s 

/ 

/ 

/ 

y  '  / 

/ 

/ 

/ 

«$ 

/ 

/ 

0 

^ 

*A 

7 

/' 

< 

\y 

7 

/ 

z 

^ 

50      100 


150   200  250  30O  3SO  4OO 

Volbs 


Fig.  171. 

the  current  is  2'2  amperes  and  the  power  125  watts  at  the  normal 
terminal  pressure  of  400  volts.  Hence  the  power  component  is 
0'32  amperes,  and  the  wattless  one  is  given  by 

V(2-2)2-(0-32)2  or  2'18  amperes. 
Again,  the  expression  for  the  torque  (no.  4)  is 

£*  ?    p  «,   TT 
P 


27rn  ((JB,  +  2^')  +  2s  (S,  +  S0?} ' 


216  ALTERNATING   CURRENTS 

all  the  quantities  in  which  have  been  found   except  P  and  n 
H  is  the  number  of  pairs  of  poles  or  4,  and  n  is  the   number  of 
periods  per  second  ;  the  result  is  in  Joule-radian  units  ;  it  gives 

160,000  x  0-092  x  5*62  x  2  x  4 

2-7T  x  60  x  37-6 
=  47  Joule-radian  units  =  35  foot-pound  units. 

The  current  taken  has  the  power  component  given  in  equation 
(8)  or 


and  the  wattless  component  <$q. 
For  the  former  we  have 

^,  =  0-31 

=  0-092x400  =  36- 


2  +  2  .  2  .  R!  .  R,  =  V(5'62)2  +  2  x  0-092  x  5'62  x  3'79 

=  \/31'6  +  3'8  =  5-97. 
Hence  the  power  component  is 

0-31  +  or  6-42  amperes, 


while  the  value  of  ^  is  2'  18  amperes. 
Hence  the  current  per  phase  is 

V(2'182)  +  (G'42)2  =  6'95  amperes 
and  the  total  current  is  very  nearly  14  amperes. 

In  order  to  find  the  efficiency  at  the  output  of  4  kilowatts  we 
must  first  find  the  various  losses.  These  are  three  in  number, 
(1)  the  constant  rotational  loss  of  250  watts,  (2)  the  ohmic  loss  in 
the  stator  which  is  given  by  6'922  x  3'79  watts  per  phase  or 
364  watts  for  the  two  phases,  and  (3)  the  ohmic  loss  in  the  rotor  ; 
it  must  be  noted  that  the  rotor  current  is  the  power  component  of 
the  stator  current,  since  the  other  parts  are  concerned  with  the 
stator  only,  hence  the  ohmic  loss  in  the  rotor  is  per  phase 
5'952  x  5'62  watts  or  400  watts  for  the  two  phases.  Hence  the 
total  loss  will  be  1030  watts,  and  hence  the  input  must  be 
5030  kilowatts.  Thus  the  efficiency  is  about  80°/o- 

The  power  factor  can  be  readily  deduced,  for  the  power 
component  of  the  current  is  5  '95  amperes  per  phase  and  the 
total  current  per  phase  is  6'95,  hence  the  power  factor  is 

W--V- 

If  different  values  of  the  slip  are  assumed  and  the  above 
calculations  made  for  each  it  is  evident  that  a  table  can  be  drawn 


EQUATIONS   FOR   INDUCTION   MOTOR 


217 


up  for  the  different  quantities,  and  hence  the  relation  of  these  to 
one  selected  quantity  can  be  shown  by  curves.  It  is  usual  in 
practice  to  take  the  torque  exerted  in  foot-pounds  as  the  in- 
dependent variable  for  this  purpose,  and  the  set  of  curves  deduced 
from  the  given  curves  in  Figs.  170  and  171  for  the  motor  we  have 
been  considering  are  shown  in  Fig.  172. 

The  starting  torque  will  be  given  by  equation  (5),  being 


27m./2 

where  /  is  the  stand-still  impedance  or  181.     On  substituting 
and  reducing  this  comes  to  46  foot-pound  units. 


600- 


•2-S- 


W         15        20       25        30       35       40        45        £C 
Torque  in  Foot  Pound  Units. 

Fig.  172. 

The  slip  which  exists  when  the    motor  is  on  the   point   of 
stopping  is  given  by  equation  (4),  which  leads  to 

5-62 

°r  *  = 


On  substituting  this  in  the  equation  for  the  torque  we  get  for  the 
value  of  this  the  amount 


since  (R,  + 


0)2  +  Sf 


160,000  x  0-35  x  5-62  x  2  x  4 


2vr  x  (50  x  78 
+  S0)2  =  {5'62  +  (0'35  x  3'79))2 

+  (0-1-24  x  237-5)  or  48'5  +  29'5  =  78. 

This  gives  P  about  64  foot-pound  units  or  nearly  twice  the  full 
load  torque. 


CHAPTER  XVII. 

MULTIPOLAR   MOTORS;    STARTING;    TESTING. 

UP  to  the  present  the  case  we  have  been  considering  is  such 
that  the  rotor's  speed  is  always  nearly  the  same  as  that  of  the 
rotating  field  under  working  conditions,  or  the  rotor  makes  nearly 
the  same  number  of  revolutions  per  second  as  the  alternations.  In 
many  cases  this  is  a  much  higher  speed  than  is  either  desirable  or 
necessary  and  we  must  now  see  how  this  difficulty  is  overcome. 
The  state  of  flux  in  our  motor  has  consisted  in  the  production  of 
two  bands  of  magnetic  flux,  the  one  passing  into  the  rotor,  the 


A 

999 


B    \   A    \    B    \   A    \   B    \  A    \   B    \    A    \   B    \  A    \  B 
oo  o !  ®C)  ®!o  o  Q!®  ®  g>!  po  p;  ®  ®  ®!ooo  j  &  a  ®io  o  pi  ••»  [poo 

tttt         imi         tVttt        4411*          VVYfrV        A  *A  ^ 


e-- 

Fig.  173. 

other  exactly  opposite  to  it  and  passing  out  of  it.  Let  a  winding 
be  adopted  which  will  produce  more  than  these  two  fluxes,  for 
example  such  a  one  as  is  shown  in  Fig.  173,  where  there  will  be  six 
such  fluxes,  three  into  and  three  out  of  the  rotor.  The  winding  is 
taken  as  two-phase  and  the  circuits  fed  by  the  two  phases  are 
marked  A  and  B.  Each  belt  of  flux  will  occupy  a  certain  amount 
of  the  circumference  of  the  stator ;  if  there  be  II  pairs  of  these 
fluxes,  or  II  pairs  of  travelling  poles,  the  space  occupied  by  any 
one  band,  being  due  to  the  action  of  two  of  the  windings,  will 

occupy  an  angle  at  the  centre  of  the  amount  ®  =  -^ .     The  joint 

action  of  the  coils  will  produce  a  definite  flux  in  the  air-gap,  the 
distribution  of  this  at  any  instant  will  have  some  definite  law. 


MULTIPOLAR   MOTORS;    STARTING;    TESTING  219 

With  the  same  assumptions  as  in  the  case  of  the  two-pole  stator, 
namely  sine  distribution  in  space  along  the  angle  &  and  sine 
variation  with  time,  together  with  a  distribution  of  windings  so 
arranged  (as  shown)  that  the  two  sets  of  windings  must  have  the 
position  of  their  maxima  differing  by  the  half  of  the  angle  <*),  the 
supplied  currents  being  in  quadrature,  it  is  readily  seen  that  the 
expressions  for  the  two  fluxes  due  to  the  two  sets  of  coils  will  be 

4>  sin  ^~- 0  sin ^>£  and  <&  cos -^  Q  cos  pt 
vy  \y 

for  each  flux  distribution.  Thus  the  sum  of  these  gives  the 
actual  existing  flux.  This  leads  to  3> cos  ( -7-  6  —  pt\  as  repre- 
senting the  flux  at  any  point  specified  by  6  and  any  time 
specified  by  t ;  this  means,  by  the  same  reasoning  as  before,  that 
the  angular  velocity  with  which  the  band  of  flux  rotates  is 

~—  p  or  ^ .     Thus  for  each  pair  of  windings  in  one  phase  there 

will  be  a  belt  of  flux  rotating  at  the  above  velocity,  which  can 
be  given  widely  different  values  for  any  definite  periodic  time  by 
altering  the  number  of  pairs  of  poles,  II.  It  can  easily  be  shown 
that  the  similar  result  holds  for  a  three-phase  winding  of  more 
than  two  poles  per  phase. 

Starting  apparatus.  In  considering  the  conditions  at 
starting  we  saw  that  there  was  considerable  advantage  gained, 
both  in  torque  and  in  lessening  the  current  required,  if  the  rotor 
had  a  higher  resistance  than  in  its  normal  running  state.  If  the 
winding  of  the  rotor  be  made  in  the  way  hitherto  supposed,  with 
bars  connected  at  the  two  ends  in  a  permanent  manner,  it  is  evident 
that  it  is  not  possible  to  alter  at  will  the  resistance  of  the  rotor. 
In  order  to  be  able  to  do  this  it  is  necessary  to  wind  the 
rotor  in  some  way  so  as  to  permit  of  the  insertion  of  resistance. 
Any  form  of  winding  will  in  general  be  suitable,  but  in  practice  it 
is  found  best  to  employ  one  formed  of  three  sets  of  coils  wound 
much  in  the  form  of  an  ordinary  three-phase  winding.  The  one 
point  that  must  be  kept  in  view  in  settling  this  winding  is  to 
avoid  joining  in  series  wires  which  have  opposing  E.M.F.S  being 
generated  in  them.  Thus  if  the  stator  is  so  wound  as  to  produce 
an  odd  number  of  pairs  of  poles  the  rotor  may  be  wound  with 
circuits  that  are  diametral,  since  in  this  case  it  is  evident  that 
the  wires  so  joined  will  be  in  opposite  fields  at  the  same  moment; 
but  if  the  poles  be  even  in  number  this  would  result  in  the  two 
opposite  wires  of  any  coil  on  the  rotor  cutting  the  rotating  fluxes 
in  the  same  direction,  and  hence  in  this  case  the  winding  across 
the  diameter  would  be  inadmissible.  Since  it  is  desirable  to  place 
resistances  in  all  three  circuits  on  the  rotor,  a  star  connection  of  its 
windings  will  be  required.  The  free  ends  of  this  star  winding  will 


220  ALTERNATING   CURRENTS 

be  brought  to  three  rings  on  the  shaft  such  as  is  shown  in  Fig.  174. 
In  many  cases  wound  rotors  are  employed  even  when  no  provision  is 
made  for  starting  resistances,  and  then  either  star  or  mesh  con- 
nections can  be  used.  In  either  case  when  the  windings  are 
short-circuited  the  effect  produced  is  practically  the  same  as  with 
the  former  bar  winding. 


Fig.  174. 

When  a  wound  rotor  provided  with  slip  rings  is  to  be  started, 
the  pressure  from  the  mains  is  applied  to  the  stator  and  the  rings 
of  the  rotor  are  attached  to  a  star-wound  resistance  which  is 
capable  of  being  cut  out  in  steps,  the  last  point  on  it  being  that  of 
short-circuit  for  the  star  winding.  In  the  cases  where  a  perma- 
nently wound  rotor  is  present  with  no  rings,  the  full  pressure 
cannot,  except  for  very  small  sizes,  be  applied  to  the  stator  until 
the  rotor  has  got  up  speed,  since  extremely  large  currents  would 
flow.  In  this  case  the  stator  has  each  phase  fed  by  an  auto- 
transformer  during  the  starting  period  so  that  only  a  fraction  of 
the  full  pressure  is  applied.  When  the  rotor  has  got  up  to  speed 
the  stator  is  put  direct  on  the  mains.  The  operations  are  performed 
by  means  of  a  set  of  double  throw  swithes  placed  on  the  top  of  a 
case  containing  the  transformers,  and  the  whole  is  called  a  starting 
box.  Since  we  have  seen  that  the  torque  produced  by  the  motor 
varies  as  the  square  of  the  applied  pressure,  only  a  small  fraction 
of  the  ordinary  torque  will  be  produced  in  this  case,  hence  this 
method  is  principally  used  to  start  up  a  motor  on  no-load. 

Efficiency,  etc.  If  a  suitable  wattmeter  be  available  the 
input  in  watts  per  phase  can  be  measured  in  either  by  using 
one  phase  only  and  assuming  a  state  of  balance  to  exist  between 
the  phases,  or  by  means  of  the  simultaneous  use  of  two  wattmeters 
as  before  described.  The  current  and  pressure  can  be  measured 


MULTIPOLAR   MOTORS;    STARTING;    TESTING 


221 


in  the  usual  way  and  thus  the  power  factor  derived.  The  output 
can  be  found  by  the  use  of  one  of  the  ordinary  types  of  brakes. 
If  the  alternations  of  the  source  of  supply  be  known,  and  the 
number  of  pairs  of  poles  on  the  stator  from  the  observations  of 
the  speed  that  must  be  taken  to  find  the  output,  we  can  readily 
derive  the  slip.  The  following  method,  due  to  Dr  Sumpner, 
enables  us  to  find  this  directly.  Let  a  small  commutator  be  fixed 
on  the  end  of  the  shaft,  having  as  many  sections  as  the  stator  has 
poles,  and  let  this  be  placed  in  series  with  the  source  of  supply 
and  an  ordinary  permanent  magnet  voltmeter.  It  is  evident  that 
if  the  slip  is  zero  this  voltmeter  will  indicate  a  steady  reading ; 
but  if  the  rotor  be  moving  more  slowly  than  the  field  is  rotating 
the  voltmeter  needle  will  oscillate  slowly  to  and  fro,  each  complete 
oscillation  taking  place  in  the  time  required  for  the  rotor  to  fall 
behind  the  belt  of  flux  a  distance  equal  to  that  occupied  by  such 
a  band.  Hence  if  the  periods  of  the  current  and  the  pairs  of  poles 
are  known  the  angular  velocity  of  the  field  is  known,  and  it 
follows  that  from  the  observation  of  the  number  of  oscillations  per 
second  made  by  the  voltmeter's  needle,  the  relative  angular 
velocity,  that  is  the  slip,  can  be  at  once  derived. 

In  Fig.  175  is  given  the  result  of  such  a  direct  test  on  a  small 
induction  motor.  The  input,  efficiency,  slip,  and  power  factor 
are  shown  in  terms  of  the  torque  in  foot-pound  units.  In  addition 


to 


. 


Synchronous  Speed  "1200  R.FM. 


8  12 

Lbs  Torque  a.t  1  Ft.  Radius. 

Fig.  175. 

are  drawn  curves  of  "  apparent  power  "  and  "  apparent  efficiency." 
The  former  is  merely  a  measure  of  the  total  current  in  both 
phases  taken  by  the  motor,  the  latter  the  ratio  of  the  output 
(reckoned  in  watts)  to  the  apparent  power. 

Use  of  commutator  method  for  small  phase  angles.     It 

will  be  recollected  that  on  p.  40  a  method  of  measuring  phase 


222  ALTERNATING  CURRENTS 

angles  was  described  depending  on  the  measurement  of  a  small 
resultant  pressure,  and  it  was  there  stated  that  the  limit  of 
application  of  the  method  lay  in  providing  suitable  low  reading 
voltmeters,  which  was  difficult  in  the  case  of  alternating  current 
ones.  The  use  of  the  commutator  just  described  in  conjunction 
with  a  calibrated  direct  current  instrument  enables  this  difficulty 
to  be  overcome.  Let  the  very  small  alternating  pressure  be 
applied  to  the  voltmeter  through  the  commutator,  and  let  the 
motor  be  run  unloaded,  then  as  we  have  seen,  the  slip  being  very 
small,  the  voltmeter's  needle  will  oscillate  slowly  to  and  fro,  and 
its  maximum  reading  will  be  the  maximum  reading  of  the  given 
small  alternating  pressure.  The  virtual  value  of  this  pressure  will 
be  definitely  related  to  this  maximum  by  a  factor  depending  on 
the  form  of  the  pressure  wave  being  2*22  for  a  sine  one. 
Hence  such  virtual  value  can  at  once  be  derived  if  the  factor 
is  known.  It  was  pointed  out  that  the  special  merit  of  the 
method  lay  in  comparing  two  small  pressures,  and  if  this  method 
of  commutation  be  employed  for  both,  the  ratio  of  the  maximum 
readings  will  be  practically  equal  to  the  ratio  of  the  virtual  values. 
For  example  in  one  case  where  the  load  in  Fig.  29  consisted  of  two 
plates  of  lead  placed  in  a  bath  of  acidulated  water  the  value  of  the 
minimum  pressure  as  commuted  was  found  to  be  0'004  .  k,  where  k 
is  the  constant  referred  to  above,  while  the  drop  down  the  small 
series  resistance  was  0'6  k,  hence  the  phase  angle  is  given  by 
sin  A  =  0*0066  or  the  power  factor  is  0*99997.  In  similar  ways  it 
is  possible  to  determine  the  phase  angle  in  such  a  case  as  that 
between  the  pressures  or  currents  in  the  two  coils  of  a  transformer. 
Results  of  such  a  test  show  that  with  careful  design  the  currents 
can  be  made  antiphased  within  1/1  Oth  of  a  degree. 

Indirect  measurement.  The  indirect  measurement  of  the 
efficiency  has  been  already  considered  in  the  chapter  on  the  no- 
load  and  stand-still  tests.  Briefly  the  method  is  as  follows  :  run 
the  motor  under  no-load  and  observe  the  current  and  power  taken 
at  normal  pressure,  let  these  be  ^0,  WQ  and  £Q.  Measure  the  stator 
resistance  RQ  and  deduce  the  nett  power  for  the  rotational  loss  W  i. 

W 
Deduce   also   the  inphase   current   ^p  —  —~  and  the  quadrature 


current  <@q  =  vX2  -I-  ^,2. 

Block  the  rotor  and  pass  full  load  current,  ^,  noting  the  power 
Ws.  Deduce  the  equivalent  total  resistance  (jR0  +  J$  and  hence 
the  rotor  resistance  El  by  the  expression  <^fs2  (720  +  j^)  =  W8. 

Assume  any  load  current  ^  ;  up  to  full  load  this  is  practically 
in  phase  with  the  rotor  pressure  and  hence  the  square  of  the  stator 
current  will  be  (^  +  ^)2  +  ^/.  This  multiplied  by  R0  is  the 
stator  loss  at  that  current.  The  corresponding  rotor  loss  is  R^.^f. 
The  sum  of  these  two  and  the  loss  Wi  is  the  total  loss  at  the 


MULTIPOLAR  MOTORS;    STARTING;    TESTING  223 

assumed   current.     The   input   being   #§(%+.&/),  the  efficiency 
readily  follows. 

Combined  test.  By  taking  advantage  of  the  fact  that  an 
induction  motor  run  above  its  synchronous  speed  acts  as  a 
dynamo,  it  is  possible  to  conduct  a  test  on  a  pair  of  similar  motors 
in  the  manner  already  considered  for  a  pair  of  alternators.  The 
motors  are  provided  with  pulleys  which  differ  just  sufficiently  in 
diameter  to  give  sufficient  difference  in  speed,  when  coupled  by  a 
thin  belt,  to  allow  the  requisite  relative  slip  required  to  take  place. 
They  are  then  both  placed  on  the  supply  mains.  It  will  follow 
that  the  more  slowly  moving  one  absorbs  power  from  those  mains 
and  acting  as  a  motor  drives  the  other  above  synchronism  as  a 
dynamo,  thus  restoring  power  to  the  mains.  The  total  losses  can 
be  measured  by  a  wattmeter  placed  in  the  supply  circuit  and  the 
load  on  the  machines  deduced  from  an  additional  reading  of 
the  power  that  is  being  delivered  by  the  dynamo  action  of  the 
more  quickly  moving  machine.  With  large  machines,  in  the 
absence  of  further  data,  the  efficiency  of  the  two  can  be  deduced 
from  the  assumption  that  each  is  working  at  the  same  efficiency 
in  the  manner  previously  described  for  two  dynamos.  With 
short-circuited  rotors  variation  of  load  can  only  be  secured  by 
altering  the  relative  diameters  of  the  pulleys,  which  is  not  in 
general  convenient  or  possible,  or  by  somewhat  altering  the 
supply  pressure.  In  the  case  of  wound  rotors  different  conditions 
of  loading  can  readily  be  provided  by  placing  resistances  in  the 
rotor  circuits  which  may  conveniently  be  ordinary  star  connected 
starting  resistances.  In  general  the  losses  in  such  extra  strips 
are  small  and  can  be  neglected,  but  in  any  case  a  correction  can 
readily  be  applied  if  the  circulating  currents  in  the  rotors  and  the 
resistances  of  the  strips  are  known. 


CHAPTER   XVIII. 


THE   MONOPHASE   MOTOR. 


IF  a  two-phase  motor  be  running  and  the  circuit  of  one  of  the 
phases  be  opened  it  will  be  found  that  the  motor  still  continues  to 
run  at  nearly  the  same  speed  as  before,  but  with  the  working 
phase  carrying  about  twice  the  current  and  taking  somewhat 
more  than  twice  the  power.  It  follows  that  a  motor  wound  in  a 
manner  similar  to  that  which  we  have  hitherto  considered  can  be 
run,  under  certain  circumstances,  from  a  single  circuit.  Consider 


Fig.  176. 

a  stator  wound  with  only  two  sets  of  windings  as  shown  in  Fig.  176, 
so  that  the  effect  of  them  would  be  to  generate  a  single  alternating 
distribution  of  flux  in  the  air-gap.  As  in  the  previous  case  we 
will  assume  that  this  flux  is  distributed  in  the  air-gap  according 
to  the  ordinates  of  a  sine  curve  and  that  the  maximum  of  the  flux 
varies  as  a  simple  harmonic  function  of  the  time.  The  flux  at  any 
point  defined  by  the  angle  6  and  the  time  t  can  then  be  written 
in  the  form  4>  sin pt .  sin  0.  But  we  can  write  this 

<I>  <E> 

—  cos  (pt  -  0)  -  y  cos  (pt  +  0), 

which  shows  that  the  single  alternating  distribution  can  be  looked 
on  as  being  equivalent  to  two  rotating  fluxes  of  the'  nature  we 
have  already  discussed,  each  of  half  the  maximum  of  the  impressed 


THE   MONOPHASE   MOTOR 


225 


alternating  flux,  that  one  of  them  will  rotate  with  the  angular 
velocity  H  =  p  in  one  direction,  while  the  other  rotates  with  the 
same  angular  velocity  in  the  opposite  direction,  that  is  with  the 
angular  velocity  H  =  —  p.  Let  the  rotor  be  revolving  with  the 
angular  velocity  CD,  then  it  will  have  a  slip  of  H  —  G>  with  regard 
to  the  flux  that  is  rotating  in  the  same  direction  as  its  own,  but 
one  of  —  (H  +  &>)  with  reference  to  the  other  rotating  flux.  With 
the  notation  used  on  p.  184  it  follows  that  the  torque  the  rotor  will 
exert  will  be  given  by 


Two  points  will  follow,  at  starting  when  &>  is  zero  the  torque  is 
zero,  as  is  evident  from  reasons  of  symmetry,  also  synchronism  will 
not  be  so  nearly  approached  as  in  the  two-phase  motor,  since  if 
&)  =  H  the  torque  is  negative.  The  mechanical  characteristic  can  be 
readily  derived  from  a  consideration  of  that  for  the  former  case  given 
on  p.  213.  In  Fig.  160  was  shown  the  complete  curve  connecting 
torque  and  speed  for  an  ordinary  two-phase  motor  from  a  value  of 
&>  equal  to  fl  to  one  equal  to  —  H.  It  will  be  recollected  that  the 
part  SB  of  this  curve  corresponds  to  dynamo  action  of  the  motor. 


Tor-c^ue 

Fig.  177. 

The  monophase  motor  will  act  as  if  it  were  operating  at  the  same 
time  on  both  parts  of  the  curve  and  thus  the  curve  for  it  will  be 
derived  by  drawing  the  part  SB  of  the  curve  in  the  reversed 
direction  and  subtracting  the  ordinates.  Thus  the  dotted  curve 
in  Fig.  177  is  the  mechanical  characteristic  of  the  monophase 
motor ;  it  will  be  seen  that  it  operates  in  a  very  similar  manner  to 
the  polyphase  form. 

Form  of  Flux  Band.  In  the  case  of  the  polyphase  motor 
with  sinusoidal  currents  we  found  that  the  rotor  current  was  nearly 
the  image  of  the  stator  band.  We  must  now  consider  the  form  of 
the  bands  in  the  present  case.  Since  the  stator  is  supplied  with 
but  a  single  current  it  will  have  a  stationary  alternating  current  band. 
L.  15 


226  ALTERNATING   CURRENTS 

Let  the  line  XXlt  Fig.  178,  be  such  as  to  have  the  position  of  this 
band,  and  a  length  such  that  OX  is  the  amplitude  of  the  same. 
Then  the  current  band  must  be  looked  on  as  being  given  by  a 
vector  along  XXl  whose  length  varies  harmonically  with  time. 
It  has  been  seen  above  that  such  a  harmonic  variation  is 
equivalent  to  two  rotating  current  bands,  each  representable  by  a 
vector  of  half  the  length,  XXlt  rotating  in  opposite  directions,  and 
coinciding  in  position  at  points  on  XX^  Let  OP  represent  either 
of  these  bands  at  the  instant  they  coincide  in  the  direction  OX. 


Fig.  178. 

Let  the  direction  of  rotation  of  the  rotor  be  contrary-clockwise,  and 
let  the  slip  relative  to  the  field  rotating  in  the  same  direction  be 
small.  With  regard  to  the  rotating  component  that  is  moving  in 
the  same  direction  as  the  rotor  we  can  draw  the  Heyland  circles 
given  at  DCO.  Hence,  as  far  as  this  component  of  our  impressed 
stator  field  is  concerned,  it  will  necessitate  the  existence  of  a  rotor 
band  of  constant  value  represented  by  the  line  PG  and  a 
magnetising  band  represented  by  OC.  Now  consider  the  stator 
component  that  is  rotating  in  the  opposite  direction,  and  form 
Heyland  circles  for  this.  If  a  be  the  slip  for  the  first  component 
and  fl  the  angular  velocity  of  either,  the  slip  between  the  rotor 
and  the  oppositely  rotating  component  will  be  211  —  cr.  Hence  if 
<j  be  small,  this  new  slip  will  be  very  large  indeed,  and  thus  the 
point  P  in  the  diagram  must  be  taken  very  near  to  D,  and  the 
diagram  will  take  the  form  of  the  smaller  one  shown  at  D&On 
where  0^  is  merely  OP  redrawn.  It  will  be  readily  seen  that 
the  rotor  band  corresponding  to  the  backwardly  rotating  stator 


THE   MONOPHASE   MOTOR  227 

band  is  given  by  PA,  and  the  magnetising  band  by  0&  where 
Cl  divides  DA  in  the  same  ratio  that  C  divides  DO.  Now  when 
P!  is  very  near  Dlt  PA  and  PA  are  almost  the  same  in  length, 
hence  not  only  will  OP  give  the  backwardly  rotating  band  of 
stator  flux,  but  also  the  corresponding  backwardly  rotating  band 
of  rotor  flux.  Hence  to  find  the  form  of  the  rotor  flux  we  can 
proceed  as  follows.  Draw  two  circles  with  their  centres  at  0  and 
with  radii  OP  and  OQ,  OP  being  the  backward  rotor  flux  and 
OQ  equal  and  parallel  to  PG,  the  forward  rotor  flux ;  divide  these 
circles  into  equal  parts,  proceeding  round  the  two  circles  in 
different  ways  as  shown  by  the  arrows.  Draw  radii  to  the  con- 
secutively numbered  points  and  find  the  resultant  of  these  lines. 
For  example,  when  both  current  bands  have  turned  through  300° 
the  radii  will  be  Oct.  and  OyS,  and  the  resultant,  giving  the 
position  and  value  of  the  maximum  of  the  rotor  current  band 
at  that  moment,  is  Oy.  It  will  be  seen  that  the  vector  giving 
the  value  and  position  of  the  maximum  rotor  current  lies  on 
the  elongated  ellipse  shown  while  the  stator  current  runs  up 
and  down  the  line  XX^  The  difference  between  this  elliptical 
band  and  the  stator  line  will  evidently  be  due  to  the  magnetising 
current  band  required  for  the  rotor  flux.  The  form  of  this  band 
in  the  present  case  will  very  nearly  be  given  by  a  circle  with 
radius  OG. 

In  the  case  figured  the  magnetising  current  is  much  larger 
than  is  ordinarily  the  case.  When  it  is  small,  it  is  evident  that 
the  magnetising  current  will  still  more  nearly  be  represented  by  a 
circle,  and  the  rotor  ellipse  will  become  flatter  and  more  nearly 
like  the  stator  line.  The  less  the  magnetising  current,  the  more 
nearly  will  the  two  bands  tend  to  have  the  same  form. 

Starting  Apparatus.  We  must  now  see  in  what  way  the 
rotor  of  the  monophase  motor  is  started  from  rest.  In  the  case  of 
very  small  motors  all  that  is  necessary  is  to  start  the  rotor  running 
in  one  direction  by  any  means.  The  slip  between  the  two  fluxes 
being  different,  a  torque  will  result  which  will  accelerate  the  rotor 
more  and  more  till  it  attains  as  nearly  as  possible  the  synchronous 
speed.  With  motors  of  even  quite  moderate  size  this  cannot  be 
done,  since  it  demands  a  very  large  current  from  the  mains.  If  in 
any  way  we  can  produce  a  rotating  field,  even  if  it  be  non-uniform, 
a  starting  torque  would  result.  Let  the  stator  be  wound  with  a 
second  set  of  wires ;  if  in  any  way  it  can  be  arranged  that  this 
winding  receives  a  current  which  is  out  of  phase  with  that  passing 
through  the  ordinary  winding,  such  a  state  of  things  will  result. 
Let  one  set  of  coils  called  the  running  coils  be  connected  directly 
to  the  mains  as  shown  in  Fig.  179,  and  let  the  other  set, 
called  the  starting  coils,  be  joined  to  them  through  a  switch 
with  a  non-inductive  resistance  in  series.  The  two  coils  have 
in  general  different  numbers  of  turns,  the  starting  ones  being 

15-2 


228 


ALTERNATING   CURRENTS 


wound  smaller  in  size  than  the  others  to  avoid  waste  of  space. 
Hence  in  the  present  case  the  circuit  containing  the  running  coils 
only  has  a  low  resistance  and  a  high  reactance,  while  the  starting 
coils'  circuit  has  a  fairly  high  resistance  and  considerably  less 
reactance.  Hence  the  currents  flowing  in  the  two  circuits  will 


Running  Coils 


Startin   Coils 


Fig.  179. 


Fig.  180. 


differ  considerably  in  phase  and  a  rotating  flux  will  be  produced. 
When  the  appropriate  speed  is  attained  the  starting  circuit 
is  broken.  Another  way  of  attaining  the  same  result  (due 
to  Mr  Heyland)  is  shown  in  Fig.  180.  Instead  of  winding  the 
starting  coils  in  slots  of  the  same  form  as  the  ordinary  ones  a 
special  form  of  slot  is  provided.  These  slots  are  large  and  are  well 
enclosed  by  iron,  and  hence  the  leakage  field  they  produce  will  be 
large,  and  if  the  resistance  be  low  the  currents  in  them  will  lag 
very  greatly  and  will  be  greatly  out  of  phase  with  the  current  in 
the  main  coils,  which  will  again  result  in  a  rotating  flux  :  the  two 
coils  are,  as  in  the  first  case,  put  in  parallel  on  the  mains,  and 
the  starting  ones  are  cut  out  when  full  speed  is  attained.  This 
method  has  the  advantage  of  occupying  very  little  of  the  useful 
working  winding  space,  and  gives  a  large  phase  difference  between 
the  two  currents. 

Such  starting  devices  can  in  all  cases  be  supplemented  by 
providing  in  addition  a  wound  rotor  provided  with  slip  rings  and 
a  resistance  as  mentioned  on  p.  219.  A  motor  with  the  Heyland 
winding  and  these  slip  rings  can  be  arranged  to  take  but  little 
more  current  in  starting  than  it  consumes  at  full  load,  a  very 
important  point  in  connection  with  regulation. 

Cascade  working.  It  will  be  seen  that  the  relation  between 
the  torque  and  speed  of  the  rotary  field  motor  bears  a  very  close 
resemblance  to  that  of  the  ordinary  shunt  motor.  In  cases  where 
it  is  necessary  to  have  a  large  starting  torque  and  a  variable 
speed  the  former  can  only  be  got  in  this  case,  as  we  have  seen,  by 
providing  the  rotor  with  an  adjustable  resistance  which  necessitates 
a  loss  of  energy.  It  is  possible  by  means  of  altering  the  number 
of  poles  of  the  motor  to  get  several  speeds  of  the  machine,  thus  if 


THE   MONOPHASE   MOTOR  229 

the  connections  of  a  four-pole  motor  are  rearranged  so  that  it  is  made 
into  a  two-pole  one,  it  will  run  at  twice  the  speed,  but  no  continuous 
speed  variation  can  be  produced  other  than  by  the  wasteful  method 
of  putting  resistance  in  the  rotor.  With  two  motors  another 
solution  is  possible :  the  second  machine  can  be  arranged  so  that 
its  stator  is  fed  from  the  rotor  currents  of  the  first,  a  method 
corresponding  in  some  respects  to  the  ordinary  practice  of  series- 
parallel  control  with  two  series  motors.  The  rotor  of  the  first 
motor,  that  on  the  mains,  must  of  course  be  wound  with  a  three- 
phase  winding  of  such  a  sort  as  to  produce  the  desired  pressure  on 
the  stator  of  the  second  one.  Such  an  arrangement  of  two 
induction  motors  is  called  a  cascade  or  concatenation  one.  Let  us 
consider  the  case  where  the  motors  are  so  connected  mechanically 
that  they  must  run  at  the  same  speed,  co,  then  if  ft  be  the  speed  of 
the  field  in  the  first  stator,  it  follows  that  the  speed  of  the  field  in 
the  second  one  will  be  II  —  &>.  Hence  when  the  second  one  is 
running  synchronously  with  its  own  field  so  that  its  slip  is  zero, 
we  must  have  O  —  &>  =  &>.  Hence  the  synchronous  speed  for  two 
motors  in  cascade  is  half  that  of  either.  Further  since  the  second 
motor  in  this  case  is  doing  no  work,  as  the  slip  is  zero,  the  first 
motor  will  be  doing  none  also,  and  thus  half  speed  will  be  the 
limiting  condition  for  no-load  in  cascade.  For  speeds  below  this 
amount,  resistance  must  be  put  in  the  rotor  of  the  second  motor,  and 
since  an  energy  current  will  then  be  flowing  into  it,  both  motors 
will  produce  a  torque.  When,  by  cutting  out  this  resistance,  the 
half  speed  is  attained,  the  second  motor  can  be  cut  out  of  circuit, 
and  the  first  motor  can  again  have  resistance  put  in  its  rotor  and 
any  speed  up  to  nearly  full  synchronism  attained.  We  thus  have 
two  speeds  attainable  with  practically  full  efficiency,  the  inter- 
mediate ones  being  procured  by  means  of  resistances,  and  the 
arrangement  is  thus  somewhat  like  the  series-parallel  one  with 
direct  current  series  motors.  It  has  an  advantage  over  this  in  the 
following  respect.  While  the  car  carrying  the  motors  is  running 
at  any  speed  above  half  synchronism,  if  the  motors  be  put  in 
cascade,  the  second  one  will  be  running  above  its  synchronous 
speed,  that  is  half  speed,  and  will  thus  in  general  be  in  such  a 
condition  as  to  be  acting  as  a  generator  and  can  return  power  to 
the  circuit  from  the  kinetic  energy  of  the  car  for  a  considerable 
range,  this  cannot  be  done  well  with  the  direct  current  arrange- 
ment. One  disadvantage  is  that  during  the  cascade  the  first 
motor  is  necessarily  working  on  what  is  the  equivalent  of  a  very 
inductive  load,  and  hence  the  current  will  be  much  out  of  phase 
with  the  pressure.  It  is  only  with  the  very  highest  class  of  motor 
that  this  is  not  a  cause  of  much  difficulty.  The  stator  must  be 
so  built  as  to  have  the  minimum  possible  leakage  field  and  hence 
the  minimum  allowable  clearance.  Again,  except  during  accelera- 
tion and  stopping,  the  second  motor  does  nothing  and  is  only  ? 
dead  weight  on  the  car. 


CHAPTER  XIX. 


METERS   OPERATING  BY   MEANS   OF  A  ROTATING  FIELD. 

SOME  forms  of  alternating  current  instruments  afford  in- 
teresting examples  of  the  application  of  the  principles  of  the 
rotating  field.  In  any  integrating  instrument  it  is  necessary 
that  there  exist  two  couples,  one  tending  to  cause  motion  of  the 
rotating  part,  the  other  tending  to  oppose  that  motion.  When 
these  two  couples  are  equal  the  rotating  part  will  move  with 
constant  velocity.  In  by  far  the  great  majority  of  cases  the 
opposing  couple  is  produced  by  means  of  eddy  currents  in- 
duced in  the  moving  part,  which  generally  consists  of  a  metal 
disc  or  cylinder.  Such  eddy  currents  are  due  to  the  rotation 
taking  place  in  a  field  due  to  permanent  magnets,  and  hence  the 
retarding  couple  is  proportional  to  the  speed  of  revolution  of  the 


Fig.  181. 

disc.  In  order  that  the  total  work  registered  by  the  meter  may 
be  proportional  to  the  number  of  turns  the  disc  makes,  and  may 
therefore  be  measured  by  means  of  an  ordinary  counting  mechanism, 
it  follows  that  the  moving  couple  must  be  proportional  to  the 
power  supplied  to  the  circuit  to  which  the  meter  is  attached. 
Consider  the  apparatus  shown  in  Fig.  181,  which  consists  of  a 
set  of  stampings  provided  with  four  projecting  poles,  and  a 


INDUCTION   METERS  231 

copper  cylinder  carefully  pivoted  coaxially  with  them.  Let  two- 
of  the  poles,  A  and  A,  be  wound  with  a  few  turns  of  thick 
wire  and  placed  in  series  with  the  circuit,  L,  in  which  the  power 
is  required  to  be  measured.  Let  the  other  two  poles,  B  and  B, 
be  wound  with  fine  wire,  and  let  this  winding  be  arranged  in 
circuit  with  some  inductive  device  so  that  the  field  produced  by 
these  coils  is  in  exact  quadrature  with  that  due  to  the  series  coils 
when  the  load  is  non-inductive.  Methods  by  which  this  condition 
can  be  attained  will  be  referred  to  later  on.  Then  the  assemblage 
of  poles,  etc.,  practically  constitute  a  simple  form  of  two-phase 
motor.  We  can  show  that  the  couple  produced  by  this  arrange- 
ment on  the  pivoted  cylinder  is  very  closely  given  by  the  expression 
£ .  ^ .  cos  \,  where  £  is  the  pressure  at  the  terminals  of  the  shunt 
coils,  ^  is  the  current  in  the  main,  and  \  is  the  angle  of  lag 
between  the  two,  in  other  words,  the  applied  couple  is  proportional 
to  the  power  taken  by  the  load  attached  to  the  mains,  and  hence 
the  condition  for  operation  of  the  meter  is  fulfilled. 

Expression  for  the  torque.  We  will  assume,  as  in  the 
two-phase  motor,  that  each  flux  is  distributed  round  the  gap 
at  any  instant  in  a  sinusoidal  manner,  and  from  the  manner 
in  which  the  poles  are  placed,  if  one  varies  as  the  sine  of  the 
angular  position,  the  other  will  vary  as  the  cosine  of  the  same 
angle.  We  will  also  suppose  that  the  value  of  the  fields  at 
the  mid  points  of  the  two  sets  of  fluxes  also  varies  in  a  simple 
harmonic  manner  with  the  time.  Consider  first  that  the  circuit 
to  which  the  meter  is  attached  is  non-inductive.  The  main 
current,  passing  round  the  coils  AA,  will  produce  in  them  an* 
alternating  flux  which  will  be  proportional  to  the  current  since  the 
principal  part  of  the  magnetic  circuit  is  air,  and  will  lag  slightly 
in  time  by  the  small  angle  of  hysteretic  lead  referred  to  on  p.  48^ 
If  the  shunt  coils  are  merely  joined  by  means  of  an  ordinary- 
resistance,  in  the  same  way,  the  current  in  the  coils  BB  will  be 
proportional  to  the  pressure  and  will  lag  after  the  pressure  by  an 
angle  dependent  on  the  relative  values  of  the  resistance  and  the 
self-induction,  the  field  resulting  will  (as  before)  again  lag  a  little 
more,  due  to  the  hysteresis.  If  instead  of  simply  connecting  the 
coils  by  a  resistance  some  inductive  device  be  put  in  series,  the 
field  produced  by  the  current  due  to  the  pressure  will  still  be 
proportional  to  that  pressure,  but  the  phase  angle  can  be  so 
arranged  that  this  shunt  field  is  exactly  in  quadrature  in  time 
with  that  due  to  the  series  coils.  Hence  if  we  assume  that  the 
series  current  is  distributed  in  space  as  the  sine  of  the  position 
angle,  0,  round  the  gap  and  that  its  field  varies  in  a  simple 
harmonic  manner  with  the  time,  we  can  write  it,  <J>a  sin  pt .  sin  #, 
where  4>a  is  its  maximum  value,  which  is  proportional  to  the 
current  flowing.  In  the  case  we  are  considering,  that  of  non- 
inductive  load,  the  flux  due  to  the  shunt  coils  will  of  necessity  be 


232  ALTERNATING   CURRENTS 

distributed  in  space  according  to  the  cosine  of  0,  but  instead  of 
varying  as  the  sine  of  the  time  angle,  the  inductive  device  is  so 
arranged  that  the  phase  is  altered  so  that  it  varies  as  the  cosine  of 
the  time,  hence  it  can  be  expressed  by  <J>6  cos  pt .  cos  0,  where  <J>6  is 
proportional  to  the  pressure  on  the  terminals.  The  nature  of  the 
inductive  devices  will  be  gone  into  later  on. 

Now  let  the  load  be  inductive  so  that  the  current  lags  on  the 
pressure  by  the  angle  X;  it  is  now  evident  that  the  expressions 
for  the  two  fields  at  any  angle  0  and  any  time  t  will  be  given  by 
<E>a  sin  (pt  —  X)  sin  0  and  <l>6  cos  pt .  cos  0. 

The  working  out  of  these  expressions  is  simplified  if  we  write  for 
X  the  angle  2 a,  and  since  the  instant  from  which  time  is  reckoned 
is  of  no  moment,  the  above  can  be  written  in  the  forms 

4>a  sin  (pt  —  a)  sin  0   and   <£6  cos  (pt  +  a)  cos  0. 

Hence  the  field,  <l>,  at  any  point  is  given  by  the  sum  of  the 
above.  This  can  be  reduced  as  follows : 

4>  =  <J>a  (cos  a .  sin  pt .  sin  0  —  sin  a .  cos  pt .  sin  0) 
(cos  a .  cos  pt .  cos  0  —  sin  a .  sin  pt .  cos  0) 
,cosa 


{cos  (pt  -  0)  -  cos  (pt  +  0)} 
<i>a .  sin  a 


sin  (pt  +  0)-  sin  (pt  -  6)} 


y5-  {sin  (pt  -  0)  +  sin  (pt  +  0)} 

cos  (pt  —  0)  (  ~a  ^  ^°  j  cos  a  +  cos  (pt  +  0)  f  — ^ — b  J  cos  a 

/<|>  _<j>,\    t  /^>64-<j>  \ 

+  sin  (p£  —  0)    — ^ —    sin  a  —  sin  (£>£  +  0)    — ^ —  )  sin  a. 


Thus  <i>  contains  four  terms,  each  representing  a  circular  rotating 

field,  for  two  of  these  pt  —  Q  and  for  two  pt  =  —  0,  that  is  -y-  =_p  for 

ctt 

7/3 

one  set  and  -y-  =  —  p  for  the  other  set. 

(Jit 

Hence  the  actual  combination  of  fields  is  equivalent  to  four 
rotating  circular  fields,  two  rotating   one  way,  and  having  the 


a  «-         •  ,• 

maximum  values  —  ^  -  cos  a  and  -  •=  —  sin  a,  and  two  rotating 

-  - 

in  the  opposite  way  having  the  values 

4>6  -  Oa  ,  <I>6  +  <t>a   . 

—  —^  —  -  cos  en  and  -  —  -  -  sin  a. 


INDUCTION   METERS 


233 


If  the  cylinder  be  at  rest,  so  that  the  slip  between  it  and  each 
field  has  the  value  p,  each  field  will  produce  a  torque  proportional 
to  the  square  of  the  maximum,  hence  the  forward  torque  will  be 
proportional  to 

J  {(4>a  -f  <I>&)2  cos2  a  +  (<&a  -  ^&)2  sin2  a}, 
while  the  backward  torque  will  be  proportional  to 

J  {(4>6  -  4>a)2  cos2  a  +  (4>a  +  O6)2  sin2  a}. 

The  nett  torque  is  the  difference  of  the  two  that  is  given  by 
<£a4>6  (cos2  a  -  sin2  a),  or  <J>a3>0  .  cos  2a,  or  by  <J>a3>6 .  cos  X,  since 
X  =  2a. 

But  we  have  4>a  proportional  to  VH,  and  <E>fe  proportional  to 
£,  hence  the  torque  applied  is  proportional  to  the  mean  power 
taken  by  the  load,  and  thus  the  condition  for  the  meter  to  work 
accurately  is  fulfilled. 

When  rotation  of  the  cylinder  takes  place,  the  slip  between  the 
two  fields  and  the  former  is  no  longer  the  same,  it  is  less  than  the 
angular  velocity  of  the  fields  for  one  of  them  and  greater  for  the 
other,  hence  the  couple  is  no  longer  accurately  given  by  the  above 
expression :  in  general  the  number  of  rotations  made  by  the 
cylinder  is  a  small  fraction  of  the  alternations  of  the  pressure  (that 
is  of  the  revolutions  per  second  made  by  the  fields),  and  hence  this 
error  is  but  small. 

Sliding  field  meters.  Assuming  that  it  is  possible  to  obtain 
two  such  fluxes  as  we  have  just  considered  differing  in  time-phase 
by  being  in  quadrature,  it  is  easy  to  see  that  equivalent  constructions 
can  be  arrived  at  in  which  the  poles  do  not  travel  round  a  complete 


1 

} 

Pi 

P2 

i 

Pi 

n 

i 

i 

! 

/J 

u 

j 

Fig.  182. 

revolution  for  each  alternation,  but  merely  shift  a  definite  distance 
in  the  same  time.  For  example,  consider  the  case  shown  in  Fig.  182. 
Here  the  shunt  field  is  produced  by  an  E-shaped  set  of  stampings 
being  wound  in  such  a  way  that  when  the  shunt  flux  is  a  positive 


234 


ALTERNATING   CURRENTS 


maximum  the  poles  P±  are  north  and  the  pole  P2  is  south,  whilst 
the  conditions  are  just  reversed  with  a  negative  flux.  The  series 
coils  are  not  wound  on  an  iron  core,  but  the  direction  of  winding 
is  such  that  with  a  positive  direction  of  flux  due  to  these  coils  Pa 
is  equivalent  to  a  north  pole,  while  P4  is  a  south  one  and  vice 
versa.  Hence  it  is  evident  that  the  whole  arrangement  is  equiva- 
lent to  a  portion  of  a  crown  of  poles  forming  the  winding  of  a  two- 
phase  stator,  and  hence  the  resultant  field  will  move  through  a 
distance  equal  to  that  between  the  poles  PjPj  in  one  alternation 
of  the  currents  flowing,  since  from  the  position  of  the  two  fields 
the  zero  point  for  the  shunt  flux  is  the  maximum  one  for  the 
series  flux.  It  follows  that  the  condition  of  affairs  with  respect 
to  a  disc  pivoted  so  a&  to  be  capable  of  rotation  in  the  plane  DD 
will  be  exactly  the  same  as  that  of  our  cylinder  in  the  last  case. 
Hence,  if  an  eddy  current  brake  be  employed,  the  total  revolutions 
of  such  a  disc  will  measure  the  energy  supplied. 

Again,  consider  the  magnetic  circuit  shown  in  Fig.  183 ;  the 
top  part  is  wound  with  the  shunt  current,  the  lower  pole  with 
the  series  current.  The  flux  produced  round  the  magnetic  circuit 
by  a  current  in  the  series  coil  will  pass  across  the  gaps  gg  and 


Fig.  183. 

will  result  in  the  pole  P3  being  alternately  north  and  south. 
The  flux  due  to  the  shunt  windings  will  pass,  for  some  definite 
direction  of  the  shunt  current,  in  the  manner  shown  by  the  arrows, 
partly  passing  directly  across  the  gap  G  and  partly  across  the  two 
gaps  gg.  It  will  readily  be  seen  that  the  pole  Pl  will  vary  from 
north  to  south  while  the  pole  P2  varies  from  south  to  north. 
Thus,  suppose  the  shunt  flux  to  be  at  its  maximum  in  some 
definite  direction,  the  flux  due  to  it  will  be  somewhat  as  shown 
in  Fig.  184.  If  the  relative  time-phases  of  the  shunt  and  series 
fluxes  be  properly  adjusted,  the  series  flux  will  be  zero  at  that 
moment,  and  hence  Fig.  184  will  show  the  total  distribution  of 
flux  in  the  gap,  gg,  at  that  instant.  When  the  shunt  flux  is  zero, 
the  series  one  will,  under  these  circumstances,  be  a  maximum, 
and  the  flux  in  the  gap  will  be  as  in  Fig.  185.  Again,  when  the 
flux  .due  to  the  shunt  attains  its  maximum  value  in  the  opposite 
direction,  the  gap  flux  will  be  as  in  Fig.  186,  while  lastly  when 


INDUCTION   METERS 


235 


the  series  flux  has  its  maximum  opposite  value,  it  will  be  as  in 
Fig.  187.  Hence  the  flux  shifts,  as  before,  across  the  gap  in  the 
direction  of  the  arrows  once  per  alternation.  Thus,  as  in  the  last 
case,  we  can  apply  the  principles  proved  for  the  first  case. 


Fig.  184. 


Fig.  185. 


Fig.  186. 


Fig.  187. 


The  last  two  methods  of  producing  a  sliding  flux  are  due  to 
the  Westinghouse  Company,  the  last  being  that  employed  in  their 
latest  meters ;  the  first  method  described  is  employed  by  Messrs 
Siemens  and  others. 

Inductive  devices.  We  must  now  see  what  methods  are 
adopted  to  ensure  that  the  two  fluxes  due  to  the  series  and 
shunt  circuits  shall  be  in  quadrature,  and  we  will  first  take 
the  arrangement  used  by  Messrs  Siemens  for  the  type  of  in- 
strument shown  in  Fig.  181.  Let  two  circuits  be  placed  in 
parallel  as  shown  in  Fig.  188,  and  assume  that  they  have  exactly 

L         Q         R  V 


L         Q/ 

Tnnnnnnr  ' 


Fig.  188. 


Fig.  189. 


the  same  resistance  and  self-induction.  In  Fig.  189  is  drawn  the 
corresponding  impedance  triangle,  which  is  evidently  the  same  for 
each  circuit.  It  follows  that  if  the  circuits  be  so  arranged  that 
the  resistance  is  practically  confined  to  one  portion  of  each  circuit 
and  the  self-induction  to  the  other,  the  points  Q,  Ql  of  connection 
of  each  of  the  two  portions  will  at  any  instant  be  at  the  same 
potential.  Now  arrange  the  circuits  as  shown  in  Fig.  190.  The 
vectors  will  have  the  same  lengths  and  inclinations  to  the  impressed 
pressure  line  as  before,  but  they  must  now  be  drawn  as  shown  in 
Fig.  191,  where  OP  is  the  vector  for  the  impressed  pressure  on 
either  circuit,  OL  is  that  for  the  maximum  pressure  existing  at  the 
terminals  of  the  pure  resistance  part  of  the  one  circuit  (that  is, 
is  equal  to  the  current,  <&,  carried  by  that  circuit  multiplied 
by  the  resistance,  R,  of  the  same,  or  is  &R),  while  the  vector  PL 
measures  the  quantity  plffi  for  the  inductive  part  of  the  circuit. 


236  ALTERNATING   CURRENTS 

As  regards  the  second  circuit,  the  vector  for  the  quantity 
must  be  drawn  as  at  L^P,  while  that  for  the  quantity  pL^  must  be 
drawn  as  at  QL^.  Hence  it  follows  that  the  vector  LL^  will 
represent  the  maximum  value  of  the  pressure  between  the  points 


Fig.  190.  Fig.  191. 

Q,  Qi  in  Fig.  190.  Now  whatever  be  the  values  of  the  resistances 
and  self-inductions  of  the  four  parts  of  the  circuit,  the  points  L 
and  ^  will  always  lie  on  a  circle  drawn  with  OP  as  diameter, 
hence  by  suitably  arranging  the  values  of  these  quantities,  we  can 
get  any  desired  angular  relation  between  OP  and  LL-^.  Let  the 
points  PP  be  connected  across  the  mains  and  the  points  QQi  be 
joined  to  the  terminals  of  the  shunt  winding  shown  in  Fig.  174. 
Then  we  can  so  arrange  matters  by  adjusting  one  or  more  of  the 
coils,  that  the  flux  due  to  the  current  impelled  by  the  pressure 
between  QQ-i  has  any  desired  angle  with  reference  to  the  vector 
OP  and  hence  with  reference  to  the  field  impressed  by  the  series 
coils.  In  particular  it  can  readily  be  arranged  that  these  two 
fields  are  in  quadrature  when  the  circuit  to  which  the  instrument 
is  attached  is  quite  non-inductive,  which  is  the  required  relation 
that  has  to  be  fulfilled. 

Exact  adjustment  of  quadrature.  In  the  cases  referred 
to  in  Figs.  182  and  183,  a  very  approximate  quadrature  relation 
between  the  shunt  field  and  shunt  pressure  can  be  produced  by 
a  choking  coil  action.  In  Fig.  175  the  shunt  circuit  would  be 
capable  of  producing  but  a  small  back  E.M.F.  if  it  contained  only 
the  stampings  as  shown,  and  in  this  case  an  additional  choking 
coil  with  an  air  gap  is  provided.  In  Fig.  176  the  main  flux 
passing  across  G  is  sufficient  to  enable  the  choking  action  to 
be  provided  without  such  an  auxiliary  coil.  In  both  cases,  however, 
the  quadrature  relations  will  not  be  accurately  fulfilled,  and  this 
point  must  now  be  considered. 

Take  the  case  where  the  load  is  non-inductive  and  let  0  V  and 
OC  (Fig.  198)  represent  the  pressure  and  current.  In  an  ideal 
case  the  current  OC  would  produce  a  flux  OFC  in  phase  with  it ; 
again  the  flux  crossing  the  gap  g,  (Fig.  176),  is  only  a  portion  of  the 


INDUCTION   METERS 


237 


total  flux  passing  round  the  circuit  due  to  the  shunt  winding.  Hence 
if  there  were  no  losses  in  hysteresis  in  the  iron  or  by  resistance  in 
the  shunt  winding  the  flux  across  g  would  evidently  be  in 
quadrature  with  the  pressure  0V  as  shown  at  OFV,  since  the 
induced  pressure  in  the  shunt,  which  is  given  by  OE,  is  the  only 
pressure  that  the  applied  shunt  pressure  has  to  deal  with.  In 


Fig.  192. 


Fig.  193. 


such  a  case  the  condition  of  quadrature  between  the  flux,  4>c,  due 
to  the  series  winding  and  the  flux,  4>r,  due  to  the  shunt  one  would 
be  fulfilled.  In  the  actual  case  matters  are  in  a  different  position. 
The  magnetic  circuit  is  an  iron  one,  and  hence  losses  occur  in 
hysteresis,  and  further  there  will  be  a  small  loss  of  pressure  in  the 
shunt  coil.  Consider  first  the  action  of  the  current  given  by  00 
in  Fig.  193.  As  far  as  its  connection  with  the  iron  circuit  shown 
in  Fig.  183  is  concerned,  it  forms  with  it  a  choking  coil :  on  p.  40  it 
was  seen  that  in  this  case  the  effect  of  the  hysteresis  was  to  produce 
an  angle  of  hysteretic  lead  between  the  current  and  the  flux, 
hence  the  flux  will  lag  after  the  current  by  that  angle ;  as  shown 
on  p.  51  the  presence  of  the  gap  causes  this  angle  of  lag  to  be 
reduced  to  a  greatly  smaller  value  than  would  have  been  the  case 
without  the  gap,  but  it  has  still  a  definite,  though  small,  value. 
Now  consider  the  shunt  pressure  OF  as  acting  on  the  same 
circuit  to  form  a  choking  coil.  When  resistance  is  taken  into 
account,  the  flux  produced  is  no  longer  in  quadrature  with 
the  pressure  but  inclined  at  an  angle  less  than  90°  as  shown 
at  OF'?.  Hence  in  the  case  taken,  when  the  pressure  and  current 
are  in  phase,  their  corresponding  fluxes  are  not  in  quadrature  as 
they  should  be.  Now  let  a  little  coil  of  wire  be  placed  as  shown 
at  S  on  Fig.  183,  the  ends  being  connected  by  a  short  piece  of  wire, 
so  that  the  resistance  of  the  local  circuit  thus  formed  can  be 
adjusted  at  will.  The  shunt  winding  now  forms  with  this  a  little 
transformer  instead  of  a  mere  choking  coil :  but  from  the  form  of 


238  ALTERNATING   CURRENTS 

the  iron  circuit  and  the  positions  of  the  shunt  coil  (or  primary  of 
the  transformer)  and  little  extra  coil  (or  secondary  of  the  same), 
it  is  evident  that  this  transformer  is  a  leaky  one,  and  hence  the 
flux  relations  developed  on  p.  61  will  hold.  A  reference  to  that 
page  will  make  it  clear  that  by  suitably  adjusting  the  current  in 
the  secondary,  that  is,  by  adjusting  the  resistance  in  the  little 
coil's  circuit,  we  can  cause  the  angle  between  the  pressure  vector 
for  the  applied  pressure  and  that  for  the  flux  in  the  gap,  g,  to 
have  a  considerable  range  of  values,  in  particular  this  angle  can  be 
caused  to  be  greater  than  90°.  Hence  by  this  adjustment  it  is 
easy  to  cause  the  flux  vector,  OFV,  for  the  shunt  circuit  to  lie  at 
such  an  angle  with  the  pressure  vector,  OF,  that  it  is  at  right 
angles  to  the  vector  OFC  as  shown  at  OFV.  This  adjustment  is 
extremely  simple  in  practice,  and  hence  the  method  just  described 
for  procuring  the  desired  quadrature  between  the  two  fluxes  is 
one  of  great  importance. 

This  manner  of  adjustment  for  phase  difference  can  be  applied 
to  any  form  of  meter  in  which  the  magnetic  circuits  are  such  as  to 
permit  of  the  application  of  the  little  extra  coil,  in  particular  it 
can  be  used  in  the  form  described  on  p.  233;  it  is  only  necessary  to 
place  such  a  coil  on  some  part  of  the  horizontal  part  of  the  shunt 
stamping. 

Polyphase  circuits.  In  the  case  of  polyphase  circuits  a 
meter  could  be  employed  of  the  types  just  described.  But 
another  method  is  possible.  Take  the  case  of  a  two-phase 
system,  the  current  in  one  main  is  in  quadrature  with  the 
pressure  between  the  ends  of  the  other  mains  when  the  load 
is  non-inductive.  Hence  if  we  can  arrange  matters  so  that 
the  shunt  circuit  of  the  meter  is  practically  non-inductive  and 
place  it  across  one  pair  of  mains,  and  then  place  the  series 
coil  in  one  of  the  opposite  pair  of  mains,  it  will  readily  be  seen 
that  the  required  quadrature  of  the  two  fields  will  be  very  nearly 
attained.  Hence,  provided  the  load  is  balanced,  such  a  meter 
would  indicate  correctly  for  any  power  factor.  Similarly  we  can 
utilize  the  fact  proved  on  p.  150  that  in  any  three-phase  system  with 
balanced  and  non-inductive  load,  the  current  in  any  main  is  in 
quadrature  with  the  pressure  between  the  opposite  pair  of  mains. 

Wattmeter.  Instruments  of  this  type  can  readily  be  used  as 
wattmeters.  All  that  is  necessary  is  to  provide  the  rotating  disc 
with  a  suitable  control,  such  as  a  spring;  a  pointer  fixed  to  the 
disc,  and  moving  over  a  scale,  will  then  evidently  give  a  reading  of 
the  power. 

Phase  meters.  An  interesting  application  of  the  principles 
of  the  rotating  field  is  to  a  class  of  instruments  called  phase 
meters,  the  object  of  which  is  to  show  the  phase  angle  between 


INDUCTION   METERS  239 

the  pressure  and  current  or,  preferably,  the  power  factor  of  the 
same.  Suppose  that  we  have  a  stator  of  an  induction  motor  fed 
with  alternating  polyphase  current  and  consider  for  simplicity  that 
the  phases  of  the  winding  are  fed  by  the  currents  in  the  mains, 
and  that  the  load  carried  by  those  mains  is  balanced.  Let  the 
space  where  the  motor  should  exist  be  filled  with  stampings 
simply,  and  no  winding  be  on  them.  Then  if  one  of  the  currents 
is  given  by  c  =  Csinpt  it  will  follow,  from  what  we  have  seen 
concerning  the  induction  motor,  that  there  will  be  a  rotating 
field  produced  in  the  air  gap  which  can  be  represented  by 
</>  =  <!>  cos  (pt  —  6)  when  the  currents  and  space  distributions  are 
sinusoidal,  0  being  the  space  angle  of  this  flux.  Now  let  the 
pressure  on  the  terminals  of  the  pair  of  mains  concerned  in  sending 
the  current  above  considered  be  given  by  e  =  E  sin  (pt  +  i/r) 
where  ty  is  its  phase  angle  relative  to  that  current,  and  let  this 
pressure  send  a  current  through  a  coil  pivoted  without  any  control 
in  the  air  gap  of  the  stampings,  this  coil  having  a  non-inductive 
resistance  R  in  series.  The  current  in  that  coil  will  then  be 


The  couple  that  will  be  at  any  instant  exerted  on  the  coil  will 
be  the  product  of  the  flux  into  this  current,  and  hence  the  mean 
couple  will  be  given  by 

1  E4>  [T 
T  R  Jo 
This  couple  will  vanish  when  the  integral  is  zero,  which  leads  to 

I    (sin  pt  cos  i|r  +  cos  pt  sin  i/r)  (cos  pt  cos  <f>  +  sin  pt  sin  <f>)dt  =  0. 
Jo 

Hence  we  immediately  derive 

sin  >/r  cos  (/>  +  cos  i/r  sin  (/>  =  0, 

since  the  integral  of  the  product  sin  pt  cos  pt  vanishes,  while  those 
of  sin2/>£  and  cos-pt  are  equal.  Thus  sin  (<£  +  i/r)  =  0,  that  is 
<f)  =  -  -fy.  But  since  the  coil  has  no  control  it  will  move  itself  into 
such  a  position  as  to  correspond  to  this  relation  being  fulfilled,  or 
to  one  showing  the  phase  angle  of  the  current  directly.  Should 
the  flux  not  follow  the  assumed  sinusoidal  space  distribution,  it 
will  still  be  the  case  that  the  coil  will  take  up  a  definite  position 
for  a  definite  value  of  i|r  even  though  its  position  is  not  such  as  to 
give  the  value  of  ty  directly.  In  fact  it  is  not  desirable  that  this 
should  be  the  case,  since  the  quantity  that  is  desired  to  be  indicated 
is  the  power  factor  and  not  i/r. 

Such  instruments  can  be  made  either  with  or  without  iron 
cores,  in  the  latter  case  the  current  from  the  mains  is  sent  through 
one  fixed  coil,  and  the  pivoted  coil  is  a  triple  one,  connected  to 
the  three  phases  across  the  mains. 


CHAPTER  XX. 

ARMATURE   REACTION   BY  SYNCHRONOUS   IMPEDANCE. 

In-phase  case.  The  first  point  to  be  considered  is  the  direct 
action  of  the  current  in  the  armature  on  the  impressed  flux 
produced  by  the  field  magnets.  In  Fig.  194  three  successive  crowns 
of  a  field  magnet  are  shown,  the  arrows  on  the  poles  showing  the 
direction  of  the  flux  produced.  Below  them  are  shown  three  coils 
of  an  armature  which,  for  the  sake  of  simplicity,  is  taken  as  being 
wound  in  the  concentrated  manner,  the  direction  of  flow  of  the 
current  being  shown  by  the  arrows  on  the  coils,  and  that  of  the  flux 
thereby  produced  by  the  arrows  in  the  air  gap.  If  the  rotation  be 
as  shown  by  the  large  arrow,  and  if  the  armature  be  drawn  at  the 
instant  the  E.M.F.  is  a  maximum,  it  will  occupy  the  position  shown 
in  the  figure.  Let  us  suppose  that  the  whole  circuit  on  which 
the  armature  is  working  is  devoid  altogether  of  self-induction, 
then  since  the  current  will  then  be  exactly  in  phase  with  the 
E.M.F.,  the  same  position  will  correspond  to  the  instant  at  which 
the  current  is  also  at  its  maximum.  Now  imagine  that  the 
armature  is  at  rest  in  this  position  and  that  we  send  through  it  a 
constant  current  of  the  same  amount  as  the  maximum  alternating 
one  that  is  actually  flowing,  this  current  would  tend  to  produce  a 
magnetic  flux  of  its  own  in  addition  to  that  which  the  field 
magnets  are  sending  round  the  circuit.  It  will  be  seen  that  in 
this  non-inductive  case,  the  field  that  the  armature  carries  will  at 
the  instant  of  maximum  cause  a  weakening  of  the  flux  in  half  the 
polar  face  of  each  pole  and  a  corresponding  strengthening  of  the 
other  half,  somewhat  as  is  shown  in  Fig.  194.  This  effect  is  similar 
to  the  cross-magnetising  effect  of  the  armature  of  a  direct 
current  machine,  and  unless  some  alteration  in  the  permeability 
of  the  pole  faces  takes  place  owing  to  the  alteration  in  distribution 
of  the  flux,  the  total  nett  flux  from  the  poles  will  not  alter  in 
amount  from  this  cause  but  the  flux  distribution  will  only  be 
distorted  in  shape ;  the  same  effect  must  occur  at  the  instant  the 
alternating  current  is  a  maximum. 

Now  consider  the  armature  to  be  rotating  with  the  current 
flowing,  as  it  moves  along  it  will  be  continually  carrying  a  less 


ARMATURE  REACTION  241 

and  less  current  till  at  the  moment  the  coil  is  opposite  the  pole, 
the  E.M.F.  (and  in  this  case  the  current  also)  will  be  zero,  and 
hence  no  effect  will  be  produced  on  the  flux.  From  this  moment 
the  next  following  coil  will  perform  an  exactly  similar  cycle  of 
effects,  since  it  will  be  replacing  the  first  one  from  instant  to 
instant.  It  follows  that  the  armature  current  will  cause,  or  tend 
to  cause,  a  pulsating  distortion  of  the  flux  issuing  from  the  poles 
without  materially  altering  the  total  flux  therefrom,  and  that  this 
pulsation  will  take  place  at  twice  the  period  of  the  current,  since 
the  time  for  one  coil  to  get  to  the  place  occupied  by  the  previous 
one  is  only  half  the  periodic  time  of  the  current. 

Inductive  case.  Now  let  the  armature  be  working  on  a 
circuit  of  very  large  self-induction,  so  that  the  lag  is  nearly  90°. 
The  E.M.F.  will  still  be  at  its  maximum  when  a  coil  is  between  two 
poles,  but  owing  to  the  lag  it  will  not  be  carrying  the  maximum 
current  till  it  gets  opposite  the  pole,  since  the  distance  between 


Fig.  194. 

two  poles  is  180°.  Thus  the  coil  (1)  will  still  have  its  maximum 
E.M.F.  at  the  position  shown  in  Fig.  194,  but  the  current  will 
not  be  at  its  maximum  till  the  coil  has  moved  on  to  the  position 
shown  in  Fig.  195.  As  before,  consider  the  effect  of  the  corre- 
sponding direct  current,  the  armature  being  at  rest.  It  will  be 


Fig.  195. 

seen  that  the  current  flows  in  such  a  direction  as  to  oppose  the 
magnetising  effect  of  the  field  magnet's  winding,  in  other  words  it 
tends  to  demagnetise  the  circuit.  When  the  armature  rotates, 
generating  an  alternating  current  having  this  position  in  space,  it 
will  likewise  tend  to  demagnetise,  but  the  effect  will  be  a  pulsating 
one  in  the  same  way  the  cross-magnetising  one  was,  and  it  will 
pulsate  also  with  a  periodicity  twice  that  of  the  current  itself, 
but  still  the  average  effect  of  the  alternating  current  will  on  the 
whole  be  a  demagnetising  one.  It  should  be  noted  that  the 
current  has  its  biggest  values  just  in  the  place  where  it  is  so 

L.  16 


242  ALTERNATING    CURRENTS 

situated  as  to  produce  the  most  effect  as  regards  alteration  of 
the  main  flux. 

Leading  current.  Now  let  the  current  be  one  which  leads 
the  E.M.F.  by  nearly  90°.  Instead  of  the  coil  (1)  being  opposite 
the  pole  at  the  instant  of  maximum  current  it  follows  from  what 
was  before  said  that  the  coil  (2)  will  be  in  that  position,  and  hence 
it  will  be  readily  seen  from  Fig.  196  that  a  leading  current  will 
increase  the  flux  produced  by  the  field  instead  of  diminishing  it. 


Fig.  196. 

For  lags  or  leads  between  zero  and  quadrature  the  effect  will  be 
a  combination  of  the  two  cases.  A  current  lagging  less  than  90° 
will  both  cross  magnetise  and  demagnetise  the  poles  while  a 
similarly  leading  one  will  both  cross  magnetise  and  increase  the 
flux.  We  may  say  that  the  cross-magnetising  will  be  proportional 
to  the  component  of  the  current  that  is  in  phase  with  the  E.M.F. 
and  the  demagnetising  effect  or  the  increased  magnetising  effect 
will  be  proportional  to  the  component  in  quadrature  therewith. 
These  points  must  be  borne  in  mind  when  we  come  to  consider 
the  case  of  the  synchronous  motor. 

True  Reactance  of  Armature.  The  armature  thus  pro- 
duces a  direct  effect  on  the  flux  impressed  on  it  by  the  field 
magnet,  but  in  addition  it  can  produce  a  specific  effect  on  itself 
due  solely  to  the  current  it  is  carrying  and  otherwise  independent 
to  a  large  extent  of  the  action  of  the  field  magnet.  For  when  a 
current  is  flowing  in  the  armature  it  will  produce  a  leakage  flux 
in  the  surrounding  space,  principally  through  paths  in  the  air, 
which  flux  will  be  therefore  almost  proportional  to  the  current 
flowing :  this  flux  is  the  same  in  nature  as  the  leakage  flux  of  a 
transformer  or  induction  motor,  and  can,  as  in  those  cases,  be 
looked  upon  as  endowing  the  armature  with  a  true  reactance  or 
self-induction.  This  flux  will  therefore  produce  a  definite  E.M.F. 
in  the  armature  which  will  be  in  quadrature  with  the  current 
producing  it  and  will  be  proportional  to  that  current. 

The  Synchronous  Reactance.  Since  the  armature  effect 
is  not  only  complicated  in  action  but  varies  from  instant  to  instant 
with  twice  the  period  of  the  current,  it  is  evident  that  even  an 
approximately  correct  consideration  of  the  complete  reaction  of 
the  armature,  such  as  will  be  taken  later  on,  must  lead  to  some- 
what lengthy  constructions.  For  the  purpose  of  further  elementary 
discussion  the  usual  assumption  will  at  present  be  made,  namely 


ARMATURE   REACTION 


243 


that  the  whole  effect  of  the  armature  current  can  be  taken  as  being 
representable  by  considering  it  to  possess  a  definite  constant 
reaction  in  the  ordinary  sense,  which  reactance  is  called  the 
Synchronous  Reactance  and  can  be  determined  as  follows. 

First  short-circuit  the  armature  by  an  ammeter,  run  it  at 
its  normal  speed,  or  nearly  so,  and  apply  different  currents  to  the 
field  magnets,  reading  in  each  case  the  armature  current  and  the 
field  current ;  in  this  way  a  curve  can  be  obtained  which  is 
called  the  Short-circuit  Characteristic :  the  current  taken  from 
the  armature  can  be  considerably  more  than  its  normal  full  load 
current.  Such  a  curve  for  a  small  machine  is  shown  in  Fig.  197. 
Now  let  the  armature  be  open-circuited  and  let  a  voltmeter  be 
placed  across  the  terminals  and  take,  at  about  the  same  speed, 
simultaneous  readings  of  the  exciting  current  and  the  pressure 


JZ  3  4  5  6 

Exciting  Current  in  Amperes 

Fig.  197. 

thereby  produced.  Such  a  curve  is  called  the  Open-circuit 
Characteristic  or  Saturation  Curve,  and  the  pressure  produced  at 
any  exciting  current  is  known  as  the  corresponding  Nominal 
Induced  E.M.F.  In  Fig.  197  is  drawn  so  much  of  such  a  curve  as 
corresponds  with  the  range  of  current  in  the  corresponding  short- 
circuit  test.  The  whole  curve  will  evidently  have  the  same  form 
as  the  curve  of  separate  excitation  in  a  direct  current  machine, 
that  is,  will  roughly  correspond  in  form  with  the  iron  reversal 
curve.  Such  a  complete  curve  is  given  in  Fig.  209. 

For  the  ranges  of  current  usually  employed  in  a  dynamo,  the 
short-circuit  curve  is  nearly  straight,  and  in  many  cases,  as  the 
present,  over  the  same  range  of  exciting  current  the  saturation 
curve  is  also  straight.  At  any  desired  value  of  the  exciting 
current  let  ^  be  the  value  of  the  nominal  induced  E.M.F.  and  ^ 
the  current  on  short-circuit.  Evidently  the  ratio  of  the  two  is  the 

16—2 


244  ALTERNATING   CURRENTS 

value  of  some  impedance  which  we  will  call  the  synchronous 
impedance,  or  shortly  the  impedance  of  the  armature,  this  will,  in 
the  present  case,  be  a  constant  quantity ;  let  it  be  denoted  by  /. 
Then  if  R  be  the  true  ohmic  resistance  of  the  armature  and  S  a 
quantity  called  the  synchronous  reactance  we  evidently  have 
P  =  R2  +  S'2,  from  which  S  is  readily  found. 

In  the  example  the  ratio  of  <^  to  ^  is  about  0*85  and  as  the 
resistance  was  about  one  ohm,  the  value  of  S  is  about  0*84,  the 
angle  between  the  nominal  induced  E.M.F.  and  the  current  in  the 
short-circuit  case  being  about  83J°.  In  modern  machines  of  any 
but  very  small  sizes  the  resistance  is  small  compared  with  the 
reactance,  and  hence  the  result  of  the  test  may  be  taken  to  give  S 
directly ;  in  such  case  the  angle  between  the  pressure  required  to 
force  any  current  through  the  armature  and  that  current  is  nearly 
a  right  angle.  It  follows  then  that  in  the  present  test  the  current 
can  produce  its  maximum  demagnetising  effect  since  it  lags  nearly 
90°  after  the  E.M.F.  Furthermore  the  E.M.F.  due  to  any  leakage 
field  in  the  armature  must  be  in  quadrature  with  the  current  in 
phase,  and  hence  in  this  case  both  the  demagnetising  effect  of  the 
armature  and  its  true  reactance  effect  will  be  nearly  in  phase  with 
one  another.'  The  assumption  made  that  S  is  a  constant  for  all 
conditions  of  operation  is  manifestly  untrue,  since,  as  we  have  seen, 
the  demagnetising  effect  will  be  nearly  absent  in  certain  cases,  in 
fact  the  assumption  leads  to  worse  results  than  those  actually 
found  for  a  given  dynamo. 

It  may  be  noted  that  it  is  not  necessary  that  either  of  the 
above  curves  should  be  taken  at  exactly  the  right  speed ;  in  the 
first  the  E.M.F.  at  a  given  current  will  be  proportional  to  the  speed 
and  hence  the  ordinates  can  be  easily  corrected.  In  the  second 
one  since  the  impedance  of  the  armature  is  nearly  all  due  to  the 
quantity  S  it  will  be  proportional  to  the  speed,  as  will  the 
corresponding  value  of  the  E.M.F.  acting,  hence  the  speed  need  not 
be  kept  quite  constant  at  a  given  value  throughout  the  two  tests. 

Instead  of  following  in  detail  the  exact  effect  that  the  armature 
has,  both  on  diminishing  and  distorting  the  impressed  llux  and  in 
the  possession  of  a  true  leakage  E.M.F.,  we  will  make  the  assumption 
that  the  effects  can  be  accounted  for  by  considering  that  the  machine 
possesses  (1)  a  definite  E.M.F.  due  to  a  constant  current  round  the 
fields,  which  is  called,  as  before  stated,  the  nominal  induced  E.M.F.  ; 
(2)  a  definite  and  constant  impedance  in  the  armature  consisting 
of  the  two  parts,  R  the  resistance,  and  S  the  synchronous  reactance. 
On  these  assumptions  we  shall  see  how  the  external  characteristic 
can  be  found. 

The  External  Characteristic.  Let  an  alternator  excited 
so  as  to  produce  a  nominal  induced  E.M.F.  of  the  amount  £  be 


ARMATURE   REACTION  245 

connected  to  a  circuit  such  that  the  angle  of  phase  difference 
between  the  current  in  that  circuit  and  the  terminal  pressure 
thereon  is  X. 

Let  0(7,  Fig.  198,  be  the  vector  for  any  current  flowing,  and  cut 
off  from  it  the  part  OR  equal  to  the  product  of  that  current  into  the 
armature  resistance,  this  represents  the  ohmic  drop  in  the  arma- 
ture ;  then  draw  RL  perpendicular  to  this  and  of  the  length  equal 
to  the  value  of  S  .<$,  where  S  is  the  equivalent  reactance.  The 
line  OL  will  represent  the  pressure  required  to  force  the  current 


Fig.  198. 

through  the  armature  ;  the  angle  LOR  is  a  constant  one  which  we 
will  denote  by  a.  With  a  radius  equal  to  the  nominal  induced 
E.M.F.  describe  a  circle  and  draw  the  line  OX  making  with  the 
current  the  proper  angle,  X,  for  the  phase  difference  between  the 
terminal  pressure  and  the  current  in  the  outside  circuit.  From  L 
draw  a  line  parallel  to  this  to  cut  the  circle  in  E  and  draw  EV 
parallel  to  OL.  It  is  evident  that  0  V  is  the  terminal  pressure,  £0, 
for  the  particular  current  taken.  By  proceeding  in  this  way  the 
external  characteristic  can  be  found  for  any  required  power 
factor. 

The  equation  to  the  characteristic  can  readily  be  found.     For 
from  the  triangle  OEV  we  evidently  have 


which  shows  that  the  relation  between  <^0  and  ^  is  represented  by 
an  ellipse.  For  the  purpose  of  discussion  it  is  more  convenient  to 
use  the  letter  ^  for  the  product  *2f  .  /,  so  that  &  is  proportional  to 
the  current,  and  to  consider  the  relation  between  this  quantity 
and  So.  It  is  evidently  just  the  same  in  form  as  the  external 
characteristic  since  /  has  been  taken  as  a  constant.  With  this 
notation  the  equation  to  a  characteristic  is 


246 


ALTERNATING  CURRENTS 


From  the  form  it  is  evident  that  the  origin  is  the  centre  of 
the  family  of  ellipses  represented  by  the  equation  when  the 
parameter  X  is  altered.  The  following  points  can  at  once  be  seen. 
The  axes  of  the  ellipses  are  on  the  line  at  45°  to  the  given  axes 
(Fig.  199).  Also  when  a  =  X  the  ellipse  becomes  the  straight  line 

g=£Q  +  &,  and  when  (a  —  X)  =  ~-  it  becomes  the  circle  £*  =  £*  +  ^/2. 

Further  when  cos  (a  —  X)  is  positive  the  ellipses  lie  between  the 
line  and  the  circle,  when  negative  outside  the  circle. 

In  an  ordinary  case  we  saw  that  a  is  nearly  a  right  angle,  and  in 
this  case  the  above  statements  lead  to  the  following  results  :  when 
X  is  a  right  angle,  or  the  load  is  entirely  inductive,  the  characteristic 


X 

Fig.  199. 

is  the  straight  line,  when  X  is  zero,  or  the  load  is  entirely  non- 
inductive,  the  characteristic  is  the  circle,  for  positive  values  of  X, 
that  is  for  any  inductive  load,  the  characteristic  is  one  of  the 
ellipses  lying  between  the  line  and  the  circle,  while  when  X  is 
negative,  or  the  current  leads  on  the  pressure,  the  characteristic  is 
one  of  the  outside  ellipses. 

We  thus  see  that  the  assumption  of  a  constant  equivalent 
synchronous  impedance  leads  to  practically  the  same  result  as 
that  which  we  saw  must  follow  for  the  actual  reaction  of  the 
armature,  namely  an  increased  fall  of  pressure  over  and  above  that 
incident  to  ordinary  ohmic  drop  in  the  case  of  an  inductive  load, 
and  a  possible  increase  in  pressure  when  the  load  is  a  leading  one. 
Hence  this  assumption  can  be  taken  as  giving  a  first  approxima- 
tion to  the  actual  condition  of  affairs,  and  from  its  simplicity  is  a 
useful  one  to  take  for  future  considerations. 

In  the  case  where  the  current  is  used  instead  of  the  quantity 
&  for  the  abscissae  it  is  evident  that  all  the  characteristics, 
including  that  for  non-inductive  load,  will  on  this  assumption  be 


ARMATURE   REACTION 


247 


ellipses,  with  the  sole  exception  of  that  for  a  highly  inductive- 
load,  since  the  scales  of  the  ordinates  and  abscissae  are  then 
different.  In  Fig.  200  is  given  the  true  external  characteristics  of  a 
machine  for  the  two  cases  of  non-inductive  and  highly  inductive 


Current 
Fig.  200. 

loads,  drawn  in  full  lines,  and  the  corresponding  ellipse  and 
straight  line  derived  from  the  assumption  of  a  constant  syn- 
chronous reactance  are  drawn  in  dotted.  The  errors  involved  will 
be  readily  seen.  The  full  determination  of  the  characteristic  will 
be  undertaken  in  Chap.  XXI. 

For  the  sake  of  continuity  it  will  be  desirable  at  this  point 
to  take  the  case  where  such  a  machine  as  we  have  been  con- 
sidering is  supplied  with  current  instead  of  generating  it,  or  in 
other  words  is  producing  motor  action  in  the  way  that  will  be 
more  fully  considered  in  Chap.  XXIII. 

Motor  action.  Refer  to  Fig.  194  on  p.  241  and  let  the 
armature  be  rotating  in  the  same  direction  as  before,  but  let 
the  armature  currents  be  flowing  in  the  opposite  direction  to  the 
arrows  on  the  coils.  Then  the  effect  of  the  currents  in  the 
armature  is  evidently  such  as  to  give  a  set  of  forces  which  act  in 
the  assumed  direction  of  motion,  and  the  machine  will  then  be 
operating  as  a  motor;  further,  since  it  will  be  running  at  the 
speed  corresponding  to  the  impressed  periods  of  the  pressure  on 
the  terminals,  it  is  called  a  synchronous  motor.  Owing  to  the 
armature  cutting  lines  of  force,  an  E.M.F.  will  be  generated,  and  the 
direction  of  this  E.M.F.  will  be  as  shown  by  the  original  direction 
of  the  current  arrows  on  the  armature  since  the  direction  of 
the  flux  and  of  the  rotation  are  the  same  as  before,  hence  the 
E.M.F.s  will  be  on  the  whole  in  the  opposite  directions  to  the 
currents  flowing  or  will  form  a  counter  E.M.F.  T  e  relations  are 
quite  similar  to  the  direct  current  case,  and  as  '  that  case  the 
difference  between  the  applied  pressure  and  the  -ck  E.M.F.  mu^t 


248  ALTERNATING  CURRENTS 

be  just  enough  to  force  the  current  through  the  armature,  in  this 
case,  however,  against  the  impedance  instead  of  the  resistance  of 
the  same. 

In  order  to  get  a  general  idea  of  the  effect  of  the  armature 
current  on  the  field  let  us,  as  in  the  case  of  the  dynamo,  assume 
at  first  that  the  complete  circuit  has  but  little  impedance,  then 
when  the  current  is  a  maximum,  so  will  be  both  the  pressure  and 
the  E.M.F.,  and  it  follows  that  the  armature  currents  will  only  distort 
the  field,  but  since  the  currents  are  in  the  opposite  direction  the 
distortion  is  just  opposite  to  that  in  the  dynamo.  Now  let  the 
current  lag  90°  after  the  pressure.  The  maximum  back  E.M.F.  will 
of  course  be  still  produced  when  the  armature  is  midway  between 
the  poles  as  in  the  first  figure,  but  it  will  be  seen  that  the 
armature  must  move  forward  90°  in  electrical  degrees  before  the 
current  has  attained  its  maximum  and  hence  coil  (1)  in  the  first 
figure  will  occupy  the  new  position  shown  in  the  second  figure  at 
the  moment  its  current  is  a  maximum.  But  since  the  current 
arrows  have  been  reversed,  it  will  be  seen  that  the  direction  of 
that  current  is  then  such  as  to  tend  to  increase  as  a  whole  the  flux 
in  the  field,  and  thus  in  the  motor  a  lagging  current,  that  is  one 
lagging  on  the  impressed  pressure,  increases  the  field  and  hence 
the  E.M.F.  produced  by  the  motor.  In  exactly  the  same  way  a 
consideration  of  Fig.  196  will  show  that  a  leading  current  tends  to 
demagnetise  the  field.  These  two  effects  are  thus  exactly  opposite 
to  what  occurs  in  the  dynamo. 

Just  as  in  the  case  of  the  dynamo  the  true  effect  of  the 
armature  current  is  a  complex  one,  being  partly  due  to  the  direct 
effect  of  the  current  in  it  on  the  field,  and  partly  due  to  the 
presence  of  a  true  self-induction  or  linkage  of  lines  with  its  own 
circuit  alone ;  for  rough  purposes  these  two  effects  can  be  merged 
as  before  in  a  single  term,  the  synchronous  impedance. 

Case  of  constant  load.  We  have  seen  that  the  character- 
istic of  the  dynamo  will  depend  on  the  nature  of  the  load  when 
the  nominal  induced  E.M.F.  is  a  constant,  and  it  will  be  of  interest 
to  inquire  how  the  value  of  that  E.M.F.  must  be  altered  if  it  be 
required  to  investigate  the  relation  between  it  and  the  current 
when  a  definite  load  is  to  be  supplied  at  constant  terminal  pressure, 
but  the  load  can  vary  in  regard  to  the  angle  of  phase  difference 
between  the  current  and  the  terminal  pressure.  Let  the  line  OF 
(Fig.  201)  represent  the  constant  pressure  and  let  the  load  be  taken 
>at  first  as  non-inductive.  Then  the  current  vector  will  be  in  the 
(same  direction  as  the  line  OF  and  may  be  taken  as  01.  The 
"'.mpedance  of  the  armature,  J,  will  require  a  pressure  of  the 
ajaount  ^./  to.  force  the  current  <$  through  it,  and  the  direction  of 

vector  will  -pake  with  that  of  the  current  vector  the  constant 
igle  a  above  \ferred  to,  which  is  such  that  its  tangent  is  S/R. 


ARMATURE   REACTION  249 

Let  this  line  be  VE  drawn  from  the  extremity  of  0  V.  Then  OE 
is  the  E.M.F.  that  the  dynamo  must  give.  Suppose  that  the  same 
power  is  being  delivered  but  with  the  angle  of  lag,  X :  then  the 
vector  for  this  new  current  will  be  in  the  direction  drawn  at  the 
angle  X  to  01  and  its  length  will  evidently  be  such  that  01  is  its 
projection,  hence  if  IIl  be  drawn  perpendicular  to  0V  from  /  the 
current  for  the  same  power,  but  with  the  angle  of  lag  X,  will  be 


Fig.  201. 

represented  by  the  vector  O/i.  To  determine  the  corresponding 
nominal  induced  E.M.F.  draw  the  line  VEl  making  the  same  angle, 
a,  with  the  new  current  vector  OIi  that  the  former  one  VE  did 
with  01,  and  take  the  length  of  VE^  as  being  equal  to  the 
impedance,  /,  of  the  armature  multiplied  into  the  new  current  01 \. 
Then  the  vector  OEl  will  give  the  necessary  nominal  induced  E.M.F. 
in  order  that  the  terminal  pressure  may  have  the  same  value  for 
the  same  load.  If  EE±  be  joined  we  can  easily  see  that  the  point 
E  always  moves  on  a  straight  line.  For,  from  the  construction, 
the  sides  VE  and  VEl  of  the  triangle  VEEl  are  proportional  to 
the  sides  01  and  0/j  of  the  triangle  OIIl  and  the  angle  between 
the  respective  sides  is  the  same,  hence  the  two  triangles  are 
similar,  and  thus  since  /  moves  on  the  line  IIlt  E  will  move  on  the 

line  EEl}  and  this  line  makes  with  the  line  OF  the  angle  i^  —  a. ). 

With  a  leading  current,  such  as  is  shown  by  the  line  OI2,  the  same 
construction  can  be  followed  out,  and  the  result  is  that  the  E.M.F. 
vector,  0^2,  falls  on  the  other  side  of  OE.  Hence  the  current  will 
lag  or  lead  on  the  pressure  in  this  case  depending  whether  the 
extremity  of  the  E.M.F.  vector  falls  to  the  right  or  left  of 
point  E. 


250 


ALTERNATING   CURRENTS 


It  also  follows  that  for  each  value  of  the  power  demanded  from 
the  machine  will  correspond  a  line  such  as  EE^\  this  set  of  power 
lines  can  readily  be  drawn  as  follows :  determine  the  current  *$ 
necessary  to  supply  different  powers  when  the  lag  is  zero  and  set 
off  on  the  line  VE  (Fig.  202)  distances  equal  to  the  value  of  ^ .  J. 
Through  these  points  draw  a  set  of  lines  perpendicular  to  VE  and 


Leading  Currents 

Lag&ntf  Currents 

^  

~-~\ 

/        ___ 

"\ 

/      /^ 

"s\      \ 

/  x  ^*  ^  — 

~^\£i  X     \ 

/ 

/  J?\  _ 

^7\     \     \ 

/ 

22Z  ^" 

7\     \     \     \ 

c 

2  -A^7    >^ 

^     \     \     \     \ 

z: 

^\    I    (  \ 

v    \      1      1      1      1 

B 


Line  of  Zero  Power 


Fig.  202. 

these  will  be  the  required  set  of  power  lines :  one  of  them,  namely 
that  through  V,  will  be  the  line  for  no  output.  Suppose  that  a 
definite  amount  of  power  has  to  be  supplied  as  given  by  the  power 
line  AB  and  let  the  current  be  also  given  which  has  to  supply 
this  power  at  the  constant  pressure,  the  current  vector  must  then 
lie  on  a  circle  drawn  with  its  centre  at  V  and  thus  the  required 
power  can  be  supplied  by  the  dynamo  with  a  lag,  X,  when  the 
E.M.F.  is  given  by  the  vector  OEl  and  with  the  same  lead  when 
the  vector  is  OE^.  If  various  circles  be  drawn  to  correspond  with 
various  currents  delivered  by  the  machine,  they  will  all  cut  the 
power  line  AB  in  two  points,  except  one  which  will  just  touch  it 
and  this  will  correspond  to  that  current  which  will  deliver  the 
load  when  the  circuit  is  non-inductive. 

We  can,  therefore,  determine  the  relation  between  the  main  (or 
line)  current  and  the  E.M.F.  of  the  dynamo  under  the  circumstances 
of  its  delivering  constant  power  at  constant  pressure,  but  with 
different  values  of  the  phase  angle.  As  an  example  let  it  be 
required  to  find  the  value  of  the  nominal  induced  E.M.F.  of  the 
dynamo  whose  short-circuit  curve  is  given  on  p.  243  for  a  load  of 
5  kilowatts,  when  the  phase  angle  varies  from  a  lag  of  60°  to  a 
lead  of  the  same  value,  the  terminal  pressure  being  constant  at 
100  volts. 

In  Fig.  203  set  off  100  volts  on  the  pressure  scale  and  draw  a 
line  AB  at  the  angle  a  to  this  line ;  in  this  case  a  is  an  angle 
whose  tangent  is  0'84.  When  the  current  and  pressure  are  in 
yhase  it  will  require  a  current  of  50  amperes  to  give  the  required 
load,  and  when  the  phase  angle  is  60°  the  current  will  evidently 


ARMATURE   REACTION 


251 


to  be  100  amperes.  It  will  follow  that  if  we  set  off  a  distance 
from  A  equal  to  the  product  of  the  impedance  into  the  in-phase 
current,  namely  50  x  0'84  or  42  volts,  a  circle  drawn  with 
this  as  radius  will  cut  AB  in  the  point  B  such  that  the  line 
perpendicular  to  AB  is  the  power  line  for  5000  watts.  On  the 
same  direction  set  off  pressures  such  that  each  is  the  product  of 
the  currents  given  in  the  diagram  and  the  impedance,  and  draw 


100 

^dO 

kso 

^70 
I  60 
$50 

Exciting  Current    m  Amperes 
24               6                              10             12     '        14             te              19            20            5 

\ 
\ 
\ 

\ 

/ 

,'' 

^ 

JJ 

\ 

\ 

/ 

t^' 

fs 

1 

^ 
\ 

\ 

/ 

,^ 

."' 

^ 

\ 

\ 

A 

^ 

*X^ 

X 

^^t 

^ 

•^ 

20              4O               6O              80             10 

0             12O            140             160             180           2i 

70           2 

Nominal    Induced    E.M.F.     in   Volbs. 


'°00  W^  L 


me, 


0  20  40  60 

Sca,le  of    Volts 


80 


100 


Fig.  203. 


50     60    70     80   90   100 

Amperes 


circles  to  cut  the  power  line ;  it  is  evident  that  the  distances  from 
the  origin  to  the  points  where  these  circles  cut  the  power  line  will 
give  the  two  values  of  the  E.M.F.  for  the  given  current  circle.  By 
proceeding  in  this  way  the  top  curve  connecting  the  E.M.F.  and 
current  for  the  different  definite  loads  is  easily  found. 

The  direct  observation  of  the  nominal  induced  E.M.F.  manifestly 
cannot  be  made.  But  the  complete  open  circuit  characteristic 
mentioned  on  p.  243  gives  us  the  relation  between  this  E.M.F.  and 
the  field  excitation  current,  and  hence  at  each  point  of  the  above 
curve  the  exciting  current  can  be  substituted  for  the  E.M.F.  The 
relation  thus  obtained  between  the  field  current  and  the  line 
current  is  shown  by  a  dotted  curve. 

Compounding.  It  will  thus  be  seen  that  when  an  alternator 
is  working  on  any  load,  if  the  nominal  induced  E.M.F.  due  to  a 


252  ALTERNATING   CURRENTS 

constant  excitation  be  the  sole  impressed  E.M.F.,  the  terminal 
pressure  cannot  remain  constant.  The  required  condition  of 
constant  terminal  pressure  may  evidently  be  produced  by  regula- 
tion of  the  exciting  current  in  such  a  way  as  to  produce  constant 
terminal  pressure,  the  regulation  being  either  effected  by  hand  or 
by  some  form  of  automatic  gear.  Such  a  method  is  not  the  most 
desirable  since  in  general  some  time  must  elapse  between  the 
alteration  of  terminal  pressure  and  the  adjustment  of  the  excita- 
tion. It  is  thus  necessary  to  provide  something  analogous  to  the 
compound  winding  of  a  direct  current  generator,  that  is,  in  addition 
to  the  fixed  constant  excitation  provided  by  the  ordinary  direct 
current  winding  on  the  poles,  we  must  have  a  second  source  of 
excitation  which  will  increase  with  the  demand  for  increased 
pressure  as  the  amount  and  character  of  the  load  varies.  In  an 
ideal  form  of  such  regulation  the  increased  excitation  would  be 
of  such  an  amount  as  to  produce  constant  terminal  pressure  for 
all  possible  variations  not  only  in  the  amount  of  the  load  current 
but  also  in  the  power  factor  of  the  load,  and  it  would  also  be 
instantaneous  in  its  action  so  as  to  prevent  even  momentary 
variations  in  pressure  during  the  periods  of  adjustment.  The 
latter  condition  cannot  usually  be  exactly  complied  with,  since 
even  if  the  necessary  alterations  of  excitation  be  practically  in- 
stantaneous, the  eddy  currents  in  the  field  magnets  that  will 
accompany  the  corresponding  changes  of  flux  must  of  necessity 
cause  more  or  less  delay  in  the  response  of  the  flux  to  the  alteration 
in  excitation.  Many  solutions  of  the  problem  have  been  proposed, 
but  the  following  will  serve  as  examples  of  methods  of  attaining 
the  desired  result  with  more  or  less  closeness. 

Current  Transformer.  One  very  usual  method,  which 
has  long  been  employed,  is  that  used  in  the  Westinghouse 
machines  and  is  illustrated  in  Fig.  20-4.  In  this  case  the  frame- 
work of  the  armature  carries  a  transformer  having  as  many 
primaries  as  there  are  phases ;  in  the  figure,  which  refers  to  a 
three-phase  machine,  there  are  three  such  primaries.  These  are 
put  in  series  with  the  three  armature  circuits  on  their  way  to  the 
collecting  rings.  If  the  secondary  of  such  a  transformer  were 
connected  to  any  ordinary  circuit  the  current  therein  produced 
would  evidently  be  an  alternating  current  which  would  practically 
have  its  value  equal  to  the  mean  value  of  the  current  in  the  three 
mains.  Such  a  current  would  be  of  no  utility  for  the  purpose  of 
compounding,  but  if  by  any  means  we  could  put  a  commutator  in 
series,  the  alternating  current  could  be  turned  into  a  unidirectional 
pulsating  one  as  shown  in  Fig.  205,  and  if  this  were  supplied  to 
a  second  winding  on  the  magnets,  they  would  experience  a 
pulsating  magnetising  force  which  would  be  proportional  to  the 
current  carried  by  the  armature,  and  their  magnetism  would  thus 
on  the  whole  be  increased  in  proportion  to  the  load.  Such  a 


ARMATURE   REACTION 


253 


simple  form  of  commutator  is  however  inadmissible,  for  the  setting 
of  the  commutator  would  evidently  have  to  be  very  accurate  indeed 
in  order  to  just  invert  the  connections  at  the  instant  of  zero 


11 


2OOO 


t SEPARATELY 

EXCUEO   HELD 


DIAGRAM  OF  CONNECTIONS  FOB  COMPOUND  WOUND 
THREE-PHASE  ALTERNATOR. 

Fig.  204. 

current,  further  this  zero  would  evidently  alter  with  the  power 
factor,  since  the  commutator  would  be  fixed  to  the  shaft  of  the 
machine,  and  hence  preserve  a  definite  angular  relation  to  the 
position  of  the  poles,  that  is  to  the  E.M.F.,  and  not  to  the  current. 


Fig.  205. 

To  avoid  these  difficulties,  and  also  to  give  the  possibility  of 
varying  the  amount  of  compounding  due  to  the  pulsating  series 
current,  the  commutator  is  altered  as  shown  in  Fig.  206,  which 
refers  to  a  two-pole  machine  for  the  sake  of  simplicity.  Instead 


254 


ALTERNATING   CURRENTS 


of  the  simple  reversing  commutator  that  would  be  required  for  the 
last  case,  the  commutator  itself  consists  of  two  equal  arcs  insulated 
from  each  other,  with  the  secondary  of  the  transformer  attached  to 
the  two  halves.  On  this  commutator  press  two  sets  of  brushes,  fixed 
for  any  definite  state  of  the  main  outside  circuit,  but  capable  of  being 
moved  both  as  a  whole,  and  also  as  regards  the  relative  position  of 
the  two  component  brushes  of  each  pair.  The  two  brushes  of  each 
pair  are  connected,  and  across  the  pairs  is  put  the  extra  winding 
in  which  the  transformed  pulsating  current  is  required  to  flow. 
It  will  be  seen  that  as  the  commutator  with  its  attached  secondary 
revolves  round,  the  current  from  the  latter  can  flow  round  that 
winding  while  the  angle  a  is  traversed,  but  that  both  the  winding 
and  the  secondary  are  short-circuited  while  the  angle  {3  is  being 
traversed.  Hence  it  follows  that  if  the  brushes  in  each  pair  are 


Fig.  206. 


set  exactly  alongside  so  that  a  =  180°  and  the  whole  set  is  put  in 
such  a  position  as  to  exactly  reverse  the  current  when  it  has  its 
zero  value,  the  case  is  the  same  as  the  last.  On  the  other  hand,  if  the 
brushes  in  each  pair  are  so  placed  as  to  make  a  =  zero,  or  ff—  180°, 
the  secondary  and  the  winding  will  be  always  short-circuited  and 
no  extra  effect  will  be  produced.  For  intermediate  conditions  it 
is  evident  that  it  will  be  possible  to  so  arrange  the  angles  that  the 
extra  effect  due  to  the  pulsating  current  in  the  secondary  has  any 
desired  value  between  the  maximum  one  and  zero.  The  extra 
effect  desired  evidently  depends  on  the  power  factor  of  the  load. 
For  with  a  definite  current  flowing  on  the  armature  larger  drops 
will  be  produced  with  a  low  power  factor  than  with  a  high  one,  as 
will  be  seen  from  the  consideration  of  the  characteristics  on  p.  246. 
Hence  the  relative  positions  of  the  component  brushes  must  be 


ARMATURE   REACTION 


255 


arranged  to  suit  the  load,  and  for  any  considerable  variation  of  the 
power  factor,  any  setting  that  has  been  made  will  no  longer  be 
suitable.  It  follows  that  this  system  is  principally  of  value  when 
the  load  has  fairly  constant  power  factor  or  consists  principally 
of  lighting  load  with  a  comparatively  small  proportion  of  motor 
load. 

We  have  thus  seen  that  the  alteration  of  a  will  enable  a  proper 
value  of  the  pulsating  current  to  be  provided.  But  in  general  such 
a  position  would  not  be  one  in  which  sparkless  running  was 
maintained.  In  order  that  this  may  be  the  case  it  is  evidently 
necessary  that  at  the  instant  the  commutator  either  throws  the 
coil  into  circuit,  or  short-circuits  the  coil  and  secondary,  the 
currents  should  have  the  same  values  both  in  the  circuit  of  the 
secondary  and  in  that  of  the  coil.  This  must  be  produced  by  the 
adjustment  of  the  pairs  of  brushes  as  a  whole.  The  general  nature 
of  the  necessary  conditions  is  as  follows.  During  the  active  period, 
that  corresponding  to  the  angle  a,  the  current  in  the  coil  is  directly 
under  the  influence  of  the  transformer.  During  the  angle  /?  it  is 
short-circuited  and  hence  will  gradually  fall  in  value,  being 
influenced  in  its  rate  of  fall  by  the  reaction  of  the  short-circuited 
coil  itself.  In  order  that  the  commutator  may  act  without  any 


Line  Current 


Fig.  207. 


sparking,  all  that  is  required  is  that  the  fall  of  current  during  the 
short-circuit  period  should  be  such  as  to  just  bring  the  current  to 
the  value  that  the  current  in  the  short-circuited  secondary  has  at  the 
instant  the  period  of  short-circuit  thereof  is  ended.  In  Fig.  207  are 
shown  curves  giving  the  observed  relation  between  the  angular 
motion  and  the  values  of  the  primary  and  secondary  currents, 
showing  that  by  proper  adjustment  the  required  condition  can 
be  fulfilled. 

The  compounding  is  most  satisfactory  when  a  is  considerably 
smaller  than  ft.  This  follows  from  the  fact  that  in  the  opposite 
case  the  time  the  current  has  to  fall  to  the  required  value  is  very 
small,  and  thus  any  slight  alteration  in  conditions  will  greatly 


256 


ALTERNATING    CURRENTS 


affect  its  final  value,  and  hence  the  adjustment  of  the  brushes, 
both  as  a  whole  and  relatively,  will  be  necessarily  much  more 
difficult. 

The  commutator  given  in  the  figure  is  of  the  form  suitable 
to  a  two-pole  machine,  when  the  machine  is  multipolar  the 
commutator  must  evidently  have  the  number  of  sections  increased 
to  the  proper  number,  but  the  two  pairs  of  brushes  will  still  suffice. 

Compound  pole.  The  following  method  of  securing  increase 
of  E.M.F.  with  increase  of  load  is  due  to  Mr  Miles  Walker.  On 
p.  242  it  was  shown  that  the  effect  of  the  armature  current  could  be 
considered  as  having  two  components,  the  one  producing  a  shearing 
or  distorting  effect  the  other  a  directly  increasing  or  diminishing  one : 
the  former  was  due  to  the  part  of  the  current  in  phase  with  the 


Fig.  208. 

E.M.F.  the  latter  to  that  in  quadrature.  On  circuits  of  fairly  high 
power  factor  the  former  effect  is  manifestly  the  more  important, 
and  such  cases  are  also  the  most  usual  in  practice.  The  shearing 


ARMATURE   REACTION  257 

effect  can  be  used  to  produce  a  compounding  action  in  the  follow- 
ing way.  Let  the  pole  of  the  machine  be  made  in  two  halves  as 
shown  in  Fig.  208,  the  larger  half  being  provided  with  a  magnetising 
coil  carrying  current  as  shown  and  the  smaller  half  being  devoid 
of  magnetising  current.  In  the  figure  a  second  coil  is  shown 
surrounding  the  whole  pole  round  which  current  can  be  sent  if 
desired.  When  the  armature  carries  no  load  the  distribution  of 
flux  will  be  somewhat  as  shown  at  I.  Let  the  direction  of 
running  be  such  as  to  produce  a  distribution  of  armature  current 
at  the  instant  of  maximum  in  two  adjacent  coils,  resulting  in  the 
corresponding  magnetic  effect  being  a  shear  as  shown  in  II. 
Under  the  circumstances  the  main  part  of  the  pole  is  highly 
saturated  with  the  main  flux,  and  hence  the  effect  of  the  shear  on 
it  is  comparatively  small ;  not  so,  however,  as  regards  the  small 
part  of  the  pole.  This  is  in  a  practically  non-magnetised  condition 
as  far  as  the  main  flux  is  concerned,  and  can  thus  take  up  a  large 
flux  in  response  to  the  shearing  effect  of  the  armature.  It  results 
that  the  flux  now  existing  over  the  pole  will  be  somewhat  as  in 
III,  and  thus  the  armature  current  has  produced  automatically  an 
increased  flux.  By  suitably  proportioning  the  pole  a  very  satis- 
factory compounding  effect  can  be  produced  which  is  almost  in- 
stantaneous in  its  action.  It  is  evident  that  if  a  current  with  a 
very  large  lagging  component  is  carried  the  direct  demagnetising 
effect  of  this  may  be  more  than  the  magnetising  effect  of  the 
shear,  and  then  the  compounding  will  not  occur ;  it  is  found  in 
actual  machines  that  with  values  of  the  lag  up  to  a  power  factor 
of  0*75,  which  is  smaller  than  occurs  in  most  cases,  the  com- 
pounding is  quite  satisfactory. 


17 


CHAPTER  XXL 

ARMATURE  REACTION  IN  DETAIL. 

More  detailed  consideration  of  reaction.  In  the  last 
chapter  we  developed  a  method  of  considering  the  effect  of  the 
reaction  of  an  armature  on  the  field  which  consisted  in  merging  all 
the  diverse  effects  due  to  the  different  fluxes  produced  into  a 'single 
constant  determined  from  the  open  and  short-circuit  curves,  which 
was  called  the  synchronous  reactance.  We  will  now  more  closely 
inquire  into  the  investigation  of  the  armature's  effect.  Many 
methods  have  been  proposed  but  we  will  consider  one  due  to 
Mons.  J.  Renzelman  of  Charleroi,  which  gives  excellent  results  in 
practice.  The  actual  observations  made  on  the  machine  are  few 
in  number,  and  include  the  usual  open-  and  short-circuit  curves, 
but  certain  other  constants  have  to  be  found  either  from  calcula- 
tions based  on  the  drawing  of  the  machine  or  on  experiments 
made  with  it  after  completion. 

Magnet's  stray  field.  In  the  consideration  of  the  method 
of  synchronous  impedance  one  point  in  connection  with  the  field 
magnet's  circuit  was  left  out  of  account,  the  condition  of  the 
magnetic  circuit  of  the  field  was  assumed  to  remain  constant, 
whereas  in  fact  it  does  not  do  so.  In  particular,  the  fact  that  the 
field  windings  will  produce  a  certain  flux  in  circuits  which  are 
never  cut  by  the  armature  was  left  out  of  account.  Thus  with  any 
definite  flux  in  the  armature,  which  we  will  call  the  useful  flux, 
there  will  be  associated  a  second  flux  which  passes  from  pole  to 
pole  and  never  gets  into  the  armature  at  all.  This  is  called  the 
stray  flux  of  the  field.  The  magnetic  condition  of  the  field  magnet 
is  evidently  dependent  on  the  sum  of  these  two  fluxes,  and  we 
must  first  take  this  into  account. 

Let  any  assumed  flux  be  existent  in  the  whole  armature 
circuit,  we  can  then  from  a  drawing  of  the  machine  and  a  know- 
ledge of  the  magnetic  properties  of  the  iron  of  which  it  is  formed, 
find  the  necessary  current  that  must  flow  in  the  field  coils  to 
produce  that  flux  just  as  in  the  case  of  a  direct  current  machine. 
But  the  magnetomotive  force  due  to  those  coils  will  evidently  also 


ARMATURE   REACTION   IN   DETAIL 


259 


act  on  the  paths  where  the  stray  flux  can  occur  and  will  produce  a 
corresponding  flux  therein.  The  reluctance  of  the  stray  flux  paths 
consists  mainly  of  air,  and  hence  will  be  of  practically  constant 
value,  that  is  to  say,  for  different  currents  in  the  field  coils,  the 
stray  flux  will  be  practically  proportional  to  the  current  in  the 
coils.  Let  M  be  the  reluctance  of  the  main  magnetic  circuit  for 
the  particular  current  considered,  calculated  as  indicated  above, 
and  let  D  be  the  reluctance  of  the  stray  flux  paths  calculated  also 
from  the  drawing  of  the  machine.  If  any  useful  flux  ^>  is  flowing 
in  any  particular  case,  the  corresponding  stray  flux  will  then 
evidently  be  M/D.<&. 

Now  let  OP,  Fig.  209,  be  the  complete  open-circuit  curve  for 
any  machine,  and  let  C  be  the  point  corresponding  to  that  for 
which  the  values  of  M  and  D  were  calculated.  Since  the  dynamo 
is  unloaded,  the  useful  flux  will  all  be  cut  by  the  armature,  and 
hence  the  ordinate  CM  will  be  a  measure  of  that  flux  at  any 
point.  But  for  the  particular  current  that  we  are  considering,  we 


Field  Current 


Fig.  209. 

know  that  the  stray  flux  will  bear  the  ratio  M/D  to  the  useful  one, 
hence  if  the  line  CD  be  taken  of  such  a  length  that  CD /CM  is 
equal  to  M/D,  the  length  CD  will  give  the  stray  flux  to  the  same 
scale  that  CM  gives  the  observed  useful  flux.  But  since  we  saw 
that  the  stray  flux  reluctance  was  constant,  if  the  straight  line  OD 
be  drawn,  the  true  total  flux  that  the  field  is  carrying  must  be 
reckoned  from  this  line  and  not  from  the  axis.  Thus  the  curve 
OP  can  represent  three  things;  when  the  axis  00  is  taken,  it 
represents  the  nominal  induced  E.M.F.  produced  at  any  definite 
exciting  current,  or  to  some  other  scale,  the  useful  flux,  but  when 
the  line  OD  is  taken,  the  distance  between  this  and  the  curve 
represents,  on  the  second  scale,  the  total  flux  traversing  the  field 

17—2 


260  ALTERNATING   CURRENTS 

for  the  corresponding  exciting  current.  Hence  in  any  discussion 
relative  to  the  flux  that  the  field  magnet  is  carrying,  we  must 
reckon  from  OD,  that  is,  in  any  questions  referring  to  the  saturation 
of  the  field  magnets  of  the  machine  this  line  must  be  taken  as 
axis. 

Armature's  stray  field  and  true  reactance.  In  the  case 
of  the  armature  a  similar  state  of  things  occurs.  When  any 
current  is  passing  it  produces  a  local  field  round  the  different  coils 
in  the  armature,  principally  through  air  reluctances,  which  flux 
does  not  get  across  into  the  polar  faces  at  all.  Since  the  currents 
producing  this  flux,  however,  are  alternating,  it  will  evince  its 
presence  by  the  production  of  an  E.M.F.  which  must  be  in  quadra- 
ture with  the  current  to  which  the  flux  is  due.  The  flux  is  called 
the  armature  stray  flux,  and  the  E.M.F.  can  be  called  the  E.M.F.  of 
the  armature's  stray  or  leakage  field.  Since  the  circuits  in  which 
it  flows  are,  as  said,  principally  air,  and  hence  have  constant 
reluctance,  the  flux  will  at  every  instant  be  proportional  to  the 
armature  current,  and  if  we  denote  the  value  of  the  E.M.F.  it 
would  produce  at  the  given  periodicity  for  unit  current  by  the 
letter  Ss  we  can  represent  its  effect  by  saying  that  it  is  equivalent 
to  a  reactance  in  the  armature  of  the  amount  88.  Thus,  when  any 
alternating  current  of  the  virtual  value  ^  is  flowing,  it  will  produce 
an  E.M.F.  of  the  amount  ^ .  8S  in  quadrature  with  itself. 

The  value  of  this  reactance  can,  like  that  of  the  field  magnet's 
stray  flux,  be  calculated  when  the  drawing  of  the  machine  is  given. 
It  can  be  approximately  found  in  the  same  way  as  the  reactance 
of  any  other  coil.  Let  one  phase  of  the  machine  be  selected,  if  it 
be  a  polyphase  one,  and  pass  a  current  into  it  from  a  source  of  the 
appropriate  periodicity,  and  measure  in  the  ordinary  way  the 
current,  pressure  and  power.  To  approximate  as  nearly  as  may  be 
to  the  proper  conditions,  the  armature  should  be  placed  with  its 
poles  opposite  the  spaces  between  the  field  poles ;  and  the  circuit 
of  the  latter  should  be  short-circuited  to  prevent  flux  from 
entering.  The  value  of  the  reactance  can  then  be  approximately 
determined.  As  a  rule,  the  resistance  of  the  armature  is  suffi- 
ciently small  for  the  quotient  of  the  pressure  by  the  current  to  be 
taken  as  very  approximately  the  value  of  Ss. 

Cross  and  back  reactances.     Polyphase  machine.     But 

in  the  consideration  of  the  armature  effect  in  Chap.  XX  we  saw 
that  it  also  produced  a  direct  action  on  the  field  impressed  on  it  by 
the  magnets.  In  many  ways  the  polyphase  machine  is  easier  to 
deal  with  in  this  respect,  and  we  will  first  consider  that  case.  If 
the  armature  in  such  a  case  is  carrying  a  balanced  load,  it  will  be 
roughly  equivalent  to  the  stator  of  an  induction  motor,  and  thus 
the  field  it  produces  will  be  one  which  rotates  relatively  to  the 
armature  at  a  definite  velocity  depending  on  the  impressed 


ARMATURE   REACTION   IN    DETAIL  261 

periods  and  the  number  of  poles.  If  the  windings  had  been 
spaced  out  according  to  a  sine  curve,  and  if  the  current  flowing 
were  sinusoidal,  this  field  would,  as  we  saw  in  the  case  of  the 
induction  motor,  be  one  which  varied  as  the  ordinates  of  a  sine 
curve  round  the  armature  from  pole  to  pole;  in  any  actual 
armature  it  is  not  so  shaped,  but  may  have  very  different  forms, 
but  the  field  thus  produced,  whatever  its  shape,  will  rotate, 
as  a  whole,  relative  to  the  armature.  But  this  is  itself  being 
rotated  relative  to  the  field  magnet  and  in  such  a  direction 
and  at  such  a  speed  as  to  bring  the  field  distribution  due  to  the 
armature  to  a  position  of  rest  relative  to  the  magnet's  field. 
Hence  the  action  of  the  armature  current  on  the  magnet's  field 
can  be  taken  as  being  represented  by  such  a  stationary  field,  the 
magnitude  of  which  will  depend  on  the  current's  magnitude,  and 
its  position  relative  to  the  field  will  depend  on  the  phase  of  the 
current.  We  saw  in  Chap.  XX  that  when  the  armature  was  in 
such  a  condition  that  the  whole  of  the  circuit  supplied  by  it  was 
non-inductive,  the  current  maximum  was  attained  when  the 
armature  was  exactly  midway  between  the  poles  as  shown  in 
Fig.  194,  while  with  a  completely  inductive  circuit  the  lag  was 
such  as  to  bring  the  current  opposite  to  the  poles  as  shown  in 
Fig.  195,  the  action  of  the  current  being  a  demagnetising  one  for  a 
lagging  current,  and  the  reverse  for  one  that  leads. 

Neither  of  these  positions  of  the  current  relative  to  the  field  is 
possible,  with  non-inductive  outside  load  there  must  be,  as  we  have 
seen,  the  equivalent  of  reactance  introduced  by  the  armature 
itself,  while  a  purely  inductive  load  can  never  be  attained ;  any 
actual  current  can,  however,  evidently  be  resolved  into  two 
components  having  the  two  standard  configurations  given  above. 
Thus  if  the  current  ^  have  such  a  phase  relation  as  is  shown  in 


v 


Fig.  210. 


Fig.  210  it  is  evident  that  it  is  equivalent  to  a  component  of  the 
current  given  by  <$  sin  ty  standing  opposite  to  the  poles,  and  one 
^  cos  ^  standing  between  them. 

If  the  polar  surface  were  quite  continuous  the  current  belt  in 
the  armature  would  produce  an  actual  flux  of  the  form  we  have 


262 


ALTERNATING   CURRENTS 


considered,  but  the  poles  are  discontinuous,  and  thus  the  flux  due 
even  to  an  assumed  sinusoidal  current  distribution  will  not  produce 
that  form  of  flux  in  the  poles.  Consider  Fig.  211  where  the  flux 
that  would  be  produced  is  shown  by  the  complete  curve,  which  in 
this  case  is  opposite  to  the  poles,  the  actual  flux  that  it  succeeds 


Fig.  211. 

in  getting  into  the  poles  would  evidently  be  more  nearly  repre- 
sented by  the  shaded  part.  This  can  be  evaluated  as  follows.  Let 
the  maximum  value  of  the  flux  band  be  <£  and  let  the  breadth  of 
the  pole  be  a  times  the  half-pitch  or  OCTT.  Then  reckoning  from 
the  central  line  as  zero,  the  flux  that  gets  into  the  poles  will 
evidently  be 


*/; 


cos  x  dx    or 


sm 


«7T 


Fig.  212. 

Now  consider  the  case  where  the  current  belt  is  between  the 
poles,  as  in  Fig.  212.  Here  the  expression  for  the  cross  flux 
will  be 


/2 
sin  x  dx, 
) 


which  leads  to 


ARMATURE   REACTION   IN   DETAIL  263 


Hence  the  ratio  will  be 


cross  flux  OCTT 

=  tan  -r- , 


opposing  flux  4 

which  will  depend  on  a.     With  a  =  J  it  is  0'41. 

Other  forms  of  armature  stationary  fields  would  produce 
different  values  of  this  ratio.  Take  the  extreme  case  shown  in 
Fig.  213  where  the  curve  is  flat,  it  is  evident  that  the  cross  flux 


7T—  i 


\WM\ 
J—^-i 


Fig.  213. 

and  the  opposing  ones  are  here  equal.  Again  in  the  peaked 
curve  in  Fig.  214  the  ratio  of  cross  flux  to  opposing,  as  shown 
by  the  shaded  parts,  is 


2-a' 

which  with  a  =  |  is  J.     Thus  the  ratio  of  these  fluxes  may  vary 
greatly  with  the  form  of  the  stationary  field. 


[•   (X7T    \ 


A       H\     /  \ 

Fig.  214. 

Such  fields  would  produce  E.M.F.S  in  the  armature,  the  two 
E.M.F.S  being  respectively  proportional  to  the  current  components 
concerned,  and  hence  these  may  again  be  treated  as  equivalent 
to  reactances  of  definite  amount.  If  they  are  called  the  back 
reactance,  $&,  and  the  cross  reactance,  8C,  the  corresponding 
E.M.F.S  will  be  given  by  ^ .  $& .  sin  ty  and  ^ .  Sc .  cos  i/r,  where  <$ 
is  the  current  flowing  and  ^r  is  the  angle  shown  in  Fig.  210. 

The  value  of  these  E.M.F.S  will  evidently  depend  on  the  fields 
we  have  just  been  considering  and  also  on  the  manner  in  which 
the  armature  winding  is  carried  out,  whether  with  concentrated 
or  distributed  coils.  The  calculation  of  the  same  is  in  general 
complex,  but  as  an  example  we  will  find  them  for  the  sine  distribu- 
tion of  stationary  armature  fields  that  we  have  been  taking,  namely 


264  ALTERNATING  CURRENTS 

those  shown  in  Figs.  211  and  212,  with  the  additional  assumption 
that  the  armature  windings  are  so  distributed  that  we  can  take 
the  number  of  them  per  centimetre  run  as  being  also  distributed 
according  to  a  sine  law,  that  is  the  number  of  conductors  will  be 
considered  as  being  also  proportional  to  the  ordinates  of  the  given 
flux  curves. 

Consider  first  the  case  of  the  back  flux  (Fig.  211),  here  at 
any  distance  x  from  the  centre  the  flux  will  be  <f>  cos  x  while  the 
conductors  in  a  small  length  dx  will  be,  say,  b  cos  x  dx,  thus  the 
total  E.M.F.  due  to  the  whole  set  of  conductors  under  the  poles 
will  be 


0 

which  evidently  deduces  to 

—  (a?r  +  sin  air). 

If  the  armature  had  been  able  to  produce  its  full  flux  effect 
the  E.M.F.  corresponding  would  evidently  be  given  by 


/* 

J 


2  £<|) 

cos2  x  dx   or   —  TT. 
o  ^ 


If  we  call  the  ratio  of  the  first  expression  to  this  one  Kb  the  value 
of  it  will  be 

1       . 

a  H  —  sin  OLTT 
n 

Now  take  the  case  of  the  cross  flux  (Fig.  212).  With  the 
same  assumptions  as  before  we  evidently  have  that  the  E.M.F.  due 
to  the  conductors  experiencing  that  flux  is 

r 

sin2  x  dx, 


o 

b&, 

or  is  -^-  (a-TT  —  sin  CCTT). 

If  the  whole  set  had  been  operative  the  E.M.F.  would  evidently 

b& 
have  been  as  in  the  last  case,  or  —  TT. 

Let  Kc  denote   the   ratio  of  the  above  value   to  this,  then 
we  have 

v  l     : 

Kc  =  a  --  sin  QLTT,  say. 

7T 

Thus  with  these  assumptions  the  ratio  of  these  two  quantities 
will  be 

Kc        7T«  —  Sin  OtTT  _ 
KI        7TOL  +  Sin  «7T 


ARMATURE  REACTION  IN  DETAIL  265 

The  value  of  this  ratio  can  readily  be  worked  out  for  different 
values  of  a.  When  a  is  J ,  it  will  be  found  to  be  about  0'3.  Again, 
since  the  reactances  will  evidently  be  in  proportion  to  these 

numbers  for  the  same  armature  current,  we  now  have   ~-  =  p. 

o& 

This  ratio  has  been  obtained  on  the  assumption  that  the 
reluctance  that  the  two  fields  have  to  encounter  is  the  same, 
namely  that  corresponding  to  no  flux  being  impressed  by  the 
field  magnets.  When  a  flux  exists  the  ratio  will  alter,  depending 
on  the  state  of  saturation  of  the  circuit.  As  regards  the 
back  reactance,  it  is  evident  that  this  will  be  dependent  on  the 


Fig.  215. 

field  magnet's  state  of  saturation,  but  it  can  readily  be  seen  that 
the  cross  one  will  not  be  so.  For  consider  Fig.  215,  showing  the 
position  of  the  cross  armature  field  along  a  pole  face,  any  flux 
that  this  pole  may  receive  in  addition,  due  to  its  magnetising 
current,  will  in  no  wise  alter  the  transverse  flux,  all  that  it  does  is 
to  alter  the  distribution  of  the  same  across  the  polar  face,  hence 
the  cross  reaction  is  a  constant  for  the  machine.  We  shall  see  that 
it  is  possible  to  find  the  value  of  the  back  reaction  at  one  point 
when  the  machine  is  non-saturated,  and  if  this  be  denoted  by 
JSC  it  follows  that  the  constant  cross  reactance  can  be  found, 
since  we  have  the  relation  0Sc/Sb  =  p. 

The  constancy  of  the  cross  reactance  is  of  some  importance. 
In  considering  the  question  of  parallel  running  we  shall  see  that 
the  "  synchronizing "  current  is  one  with  principally  a  power 
component,  and  hence  depends  on  this  quantity.  Such  current  is 
thus  nearly  independent  of  the  condition  of  saturation  of  the 
machine. 

Determination  of  the  reactances.  So  far,  then,  for  the 
machine  we  can  take  as  known,  the  flux  relations  given  in  Fig.  209, 
the  reactance  of  the  armature  leakage  or  stray  field,  S8,  and 
that  the  ratio  of  the  two  reactances  Sc  and  8b  is  p  =  KcjKb.  We 
must  now  see  how  we  can  find  the  absolute  values  of  these  latter 
quantities,  and  also  that  of  Sb,  the  back  reactance  in  any  state, 
saturated  or  non-saturated. 


266 


ALTERNATING   CURRENTS 


Let  the  short-circuit  characteristic  be  taken,  and  plot  it  as 
shown  at  OT  in  Fig.  216.  Then  the  curve  RQZ  can  be  drawn 
showing  the  ratio  of  the  pressure  to  the  current  in  the  test. 
Under  the  ordinary  conditions  of  an  alternator,  unless  it  is  quite 
a  small  one,  when  the  armature  is  on  short-circuit,  the  pres- 
sure produced  has  only  to  overcome  what  reactance  is  present, 
the  resistance  being  negligibly  small  in  comparison  therewith. 


Fig.  216. 

Under  such  circumstances  the  current  will  lag  practically  90°  and 
since  the  reactance  pressures  due  to  Sb  and  Ss  are  both  in  quadra- 
ture with  the  current,  they  will  under  the  circumstances  of  the  test 
be  in  phase  and  their  values  can  be  added. 

It  follows  that  the  ordinates  of  the  curve  RQZ  give  the  values 
of  (Sb  +  Ss).  But  the  value  of  Ss  is  known,  hence  the  difference 
is  the  value  of  Sb  anywhere. 

This  curve  can  also  be  used  to  find  the  value  of  Sc.  For  at  the 
origin,  the  circuit  of  the  machine  is  quite  unsaturated,  and  hence 
the  value  of  that  ordinate,  or  OR,  is  that  of  the  quantity  0Sb  +  Ss, 
where  QSb  is  the  back  reactance  for  non-saturation.  But  we  saw 
that  this  quantity  bore  the  constant  ratio  p  to  the  cross  reactance, 
hence  the  constant  value  of  that  latter  reactance  is  given  by 


and  is  thus  determined. 

If  we  could  take  the  magnetic  state  of  the  field  as   being 
the   same   under   all   circumstances,   sufficient   data  would   have 


ARMATURE   REACTION    IN   DETAIL  267 

been  obtained  to  find  the  nominal  induced  E.M.F.  corresponding  to 
any  current  and  terminal  pressure,  all  that  would  be  necessary 
would  be  to  combine  the  several  reactance  pressures  just  discussed 
with  the  ohmic  drop  in  the  armature  and  the  given  terminal 
pressure  at  its  proper  phase  relation  to  the  current,  and  the 
resultant  would  be  the  required  E.M.F. 

In  the  actual  case  things  are  different,  for  any  applied  field 
current,  which  will  be  a  constant  quantity  for  a  definite  external 
characteristic,  the  nominal  induced  E.M.F.  will  not  be  given  by  the 
corresponding  ordinate  of  the  open  circuit  curve,  but  will  depend 
on  the  alteration  in  the  magnetic  property  of  the  field  circuit 
that  is  consequent  on  the  reduction  of  the  flux  by  the  action  of 
the  armature  current.  In  Fig.  216  let  Of  be  the  constant 
exciting  current,  and  suppose  that  owing  to  the  various  reactions 
the  corresponding  nominal  induced  E.M.F.,  &,  is  given  by  the  line 
AB.  The  actual  flux  that  must  exist  with  that  current  flowing 
will  consist  of  two  parts,  that  given  by  considering  the  curve  OA  U 
as  representing  the  useful  flux,  so  that  AB  is  that  quantity,  and 
the  flux  due  to  the  current  forcing  magnetism  round  the  stray 
path  of  the  field  magnet :  the  latter  part  is/c?,  where  the  line  Od 
is  drawn  as  described  on  p.  259.  The  line  Cd  being  drawn  parallel 
to  OX  as  shown,  it  follows  that  the  total  flux  that  the  exciting 
current  is  producing  must  be  given  by  the  line  AC.  But  we  saw 
that  in  all  questions  having  reference  to  the  state  of  affairs  de- 
pending on  the  existence  of  any  definite  flux,  the  line  Od  must  be 
used  as  the  axis  and  not  OX.  Hence  if  a  point  P  be  found  such 
that  PG  is  equal  to  A  C,  all  quantities  depending  on  the  magnetic 
state  of  the  machine  must  be  taken  as  those  corresponding  to 
the  point  P.  But  the  curve  RQZ  is  that  giving  the  values  of 
(Sb  +  S8),  hence  for  the  assumed  value  of  &  the  corresponding  value 
of  (Sb  +  S8)  will  be  pQ.  It  is  evident  that  if  pq  is  the  short- 
circuit  current  corresponding  to  Pp  the  value  of  (Sb  +  S8)  is  given 
by  Pp/pq.  This  ratio  can  be  found  in  the  following  more 
convenient  manner.  Join  OP  and  produce  it  to  meet  the  per- 
pendicular from  f  in  D.  Then  it  will  be  seen  that  fD  represents 
what  the  value  of  Pp  would  have  been  with  the  exciting  current 
Of,  provided  the  magnetic  state  for  this  current  was  the  same  as 
that  for  the  point  p.  We  will  call  this  E.M.F.  <£.  Then  if  ^  is 
the  short-circuit  current  at  /,  that  is  fe,  it  is  evident  that  since 
Pp/pq  =  Dfffe,  the  value  of  (Sb  +  Sg)  for  the  actual  magnetic 
state  of  the  machine  is  given  by  £/%. 

Thus  instead  of  reading  off  the  value  of  (Sb  +  S8)  from  the 
curve,  for  any  assumed  ^,  we  can  draw  the  line  OPD  and  divide 
the  length  £^  thus  obtained  by  the  constant  quantity  ^.  It  follows 
that  by  this  construction  we  can  readily  find  all  our  three  constants 
Sb,  Ss  and  Se  for  any  assumed  value  of  £. 

External   characteristic.     Non-inductive    circuit.     We 

must  now  see  how  a  construction  can  be  developed  for  finding 


268 


ALTERNATING   CURRENTS 


the  relation  between  the  current  and  potential  difference  for  a 
constant  exciting  current.  The  case  of  a  non-inductive  load  will 
first  be  considered. 

On  referring  to  Fig.  216  it  will  be  seen  that  DF  represents  the 
difference  between  the  actual  nominal  induced  E.M.F.  and  that 
which  would  have  been  produced  with  an  ideal  magnetic  circuit 
with  the  same  exciting  current,  hence  this  difference  must  be  due 
to  the  effect  of  the  armature  current  in  forcing  back  the  flux,  in 
other  words,  this  is  the  value  of  the  E.M.F.  corresponding  to  the 
back  reactance,  8b.  Hence  if  we  draw  a  line,  OB  (Fig.  217)  to 
represent  to  some  assumed  scale  the  value  of  Sl ,  and  cut  off  the 
part  A  C  equal  to  the  corresponding  value  of  £3  the  difference  CB 
must  be  the  value  of  the  E.M.F.  due  to  the  back  reactance.  The 
E.M.F.  for  the  cross  reactance  is  in  quadrature  with  the  latter  and 
hence  will  lie  along  the  perpendicular  line  CD.  The  other  pressures 


B  F       C      G  A 

Fig.  217. 

that  have  to  be  considered  are  the  terminal  pressure,  that  for  the 
ohmic  resistance,  and  that  for  the  leakage  reactance  of  the  armature; 
the  first  two  in  this  case  are  in  phase,  the  second  is  in  quadrature 
with  them.  Thus  draw  a  line  AR  of  the  length  to  give  the 
terminal  pressure,  and  if  that  is  known,  it  will  follow  that  the 
line  RE  forming  its  production  will  represent  the  ohmic  drop,  and 
the  leakage  reactance  of  the  armature  must  lie  along  the  perpen- 
dicular, that  is  along  the  line  ED.  Hence  any  value  of  S  being 
taken,  the  corresponding  vectors  for  the  various  quantities  con- 
cerned must  form,  for  non-inductive  load,  a  figure  like  that  shown. 

Since  the  angle  at  E  is  a  right  angle,  we  can  draw  a  semicircle 
with  its  centre  on  AB  that  will  pass  through  E,  in  fact,  if  ED 
produced  cuts  AB  in  F,  the  line  AF  is  the  diameter  of  such  a 
semicircle. 

Let  a  perpendicular  EG  be  drawn  from  E  on  AB  as  shown, 
and  we  see  that  the  angles  FDC,  DEG  and  EAF  are  all  equal ; 
let  the  common  value  be  ^. 

Let  the  armature  current  flowing  have  the  value  <$,  then  since 
BC  represents  the  E.M.F.  equivalent  to  the  back  reactance  its 
value  is 

Sb.sm<*lr.(@    (1), 


ARMATURE   REACTION    IN   DETAIL  269 

also  since  DC  is  the  value  of  the  E.M.F.  equivalent  to  the  cross 
reactance  we  have 


But  from  the  figure    FC  =  DC  .  tan  -f, 
therefore  FC  =  SC.  sin  ^  .  ^. 

Thus        =    ^  is  known. 


We  can  also  write  FC  =  £b*j/-  ...........................  (2), 

06 

if  BC,  the  back  E.M.F.,  be  denoted  by  gb. 
Further  we  have         FD*  =  FC2  +  DC2, 

and  hence  FD  =  <$  .Sc.   But  from  the  meaning  of  the  symbols,  DE> 
being  the  E.M.F.  due  to  the  leakage  field  of  the  armature,  is  Sg  .  <@. 

Hence  FE  =  (Se  +  8.)  <&  ........................  (3). 

But  we  have  CG  =  DEsm^lr,  and  hence  CG  =  ($  .Sg.sm-^r,  com- 
bining this  with  (1)  we  have 

CG     Sg          nr,      *  S8 

or  OG  =        ..................  4- 


Now  AF=AC+CF,  but  since  AC  is  &  and  CF  is  given  by 
(2)  we  have 

Ar-t+Sty—d,  say    (5). 

Also  FG  =  FC+  CG,  or  from  (2)  and  (3) 

,Cf  Cf  \  /Q'lC/x 

*2»  -i-  -^  )  =  g  (   c  +    ^ 
^ft/  \     o^     / 

Hence  J.  G  being  JL  (7  —  (76r  is  given  by  (4), 

<?      P  Sg 


or 

b 

Therefore  A  F  is  divided  in  G  in  such  a  ratio  that 


It  also  follows  that         ^= 

oc  +  >« 

And  since  Sc  and  >S,  are  constant  ^  is  proportional  to 
The  terminal  pressure  is  given  by 

gQ  =  AR  =  EA-<@.r (8), 

where  r  is  the  ohmic  resistance  of  the  armature. 


270 


ALTERNATING   CURRENTS 


The  complete  predetermination  of  the  external  characteristic 
can  now  be  derived  as  follows : — 

1.  Calculate  the  ratio  of  the  useful  armature  flux  to  the  field 
dispersion  flux  for  one  field  current. 

2.  Calculate  or  measure  the  value  of  8g  for  the  armature. 

3.  Calculate  the  value  of  KC\K^  for  the  machine. 

4.  Take  an  open-circuit  characteristic. 

5.  Take  a  short-circuit  characteristic. 

From  (1)  and  (4)  we  derive  the  curve  OA  U  and  the  line  0 
(Fig.  216). 

Assume  any  voltage  for  £  less  than  the  maximum,  and  apply 
the  construction  given  on  p.  266  to  find  S^,  and  the  value  of 
($&  4-  $g)  everywhere. 

Deduce  the  value  of  $c  and  0$6  from  this  last,  (2)  and  (3). 

Deduce  the  value  of  £b  from  the  construction  on  p.  266. 

Find  the  value  of  d,  equation  (5),  draw  a  semicircle  on  a  line 
of  this  length,  and  divide  the  diameter  in  the  ratio  given  by  p, 
equation  (6). 

Erect  a  perpendicular  at  this  division  point,  then  the  current 
pressure  will  be  given  by  equation  (7)  and  the  terminal  pressure 
by  equation  (8). 

External  characteristic.  Inductive  circuit.  When  the 
external  circuit  is  inductive  the  construction  must  be  modified 


Fig.  218. 

as  follows  (Fig.  218):    with  the  same  line  as  base,  construct  a 
triangle  whose  sides  are  respectively 

-*"  tan  X   and   AK=  ^  +  ^.r.secX 


and  the  top  angle  is  (90  —  X)°  where  X  is  the  angle  of  lag.  The 
point  K  will  lie  on  a  circle  whose  centre  is  on  the  line  AO 
inclined  to  AF  at  the  angle  X.  Determine  the  point  G  as  before, 


ARMATURE    REACTION    IN    DETAIL  271 

then  if  RE  is  parallel  to  the  current  vector  and  KE  is  perpen- 
dicular, it  follows  that  since 

KR  =  f@.r.sec\,    RE  =  <@.r    and    KE  =  ritm\. 
Thus  if  we  measure  the  lengths  of  FK  and  KA  we  evidently  have 


or  is  again  proportional  to  ^Iff"  and 


When  cos  X  is  unity  this  reduces  to  the  former  expressions. 
For  a  purely  inductive  load  for  which  cosX  is  zero  and  r  is 
also  zero,  the  circle  becomes  the  diameter,  and  we  then  have 

/=  GA     and    &  = 


or,  from  the  expressions  given  above, 

BG 


The  complete  determination  of  the  external  characteristic  for 
any  assigned  power  factor  can  thus  be  carried  out.  The  method 
involves  drawing  the  given  semicircle  for  different  diameters  and 
dividing  it  in  the  proper  ratio  for  each  point,  it  is  thus  not  a  very 
rapid  one,  but  the  graphical  methods  used  can  readily  be  systema- 
tized so  as  to  save  a  good  deal  of  time.  When  applied  to  any 
machine  the  results  are  found  to  be  very  nearly  confirmed  by 
experiment,  and  are  considerably  closer  than  any  of  the  other 
methods  of  solving  the  problem. 

Monophase  machine.  We  must  now  briefly  consider  the 
case  of  the  single  phase  machine.  The  armature  field  will 
no  longer  be  one  of  uniform  strength  rotating  relative  to  the 
armature  itself  at  a  definite  speed,  and  hence  brought  to  rest 
relative  to  the  field  by  the  rotation  of  the  armature.  It  will  be  an 
alternating  field ;  but  we  saw  in  the  case  of  the  monophase  induction 
motor  that  such  a  field  could  be  considered  as  being  resolved  into 
two  oppositely  rotating  parts,  each  having  half  the  amplitude  of 
the  given  alternating  one.  The  same  holds  good  in  this  case.  When 
these  two  components  are  considered  as  being  carried  round  by 
the  rotation  of  the  armature,  it  will  be  evident  that  one  of  them 
will  be  brought  to  rest  relative  to  the  poles,  and  the  other  will  rotate 
with  double  the  angular  velocity  of  the  armature.  The  latter 
will  first  be  taken  into  consideration.  It  will  evidently  tend 
to  produce  an  alternating  current  in  the  fields  of  this  double 
frequency  as  mentioned  on  p.  241.  The  presence  of  this  current 
can  be  readily  shown  by  taking  an  observation  of  the  field  current 
by  means  of  an  oscillograph,  or  by  placing  in  the  field  circuit  two 


272  ALTERNATING   CURRENTS 

ammeters,  the  one  a  magnetic  one  which  will  only  register  the 
direct  current,  and  another  instrument,  such  as  a  hot  wire  one, 
which  will  measure  the  virtual  value.  The  latter  will  be  found  to 
show  a  greater  reading  than  the  other,  indicating  that  in  addition 
to  the  normal  exciting  current,  an  alternating  one  is  flowing.  The 
value  of  the  latter  can  evidently  readily  be  found  from  the 
readings  of  the  two  instruments.  This  pulsation  in  the  magnetising 
current  will  in  turn  tend  to  produce  higher  harmonics  in  the 
armature  which  would  again  react  on  the  field  if  circumstances 
were  favourable.  The  effect  of  these  oscillations  in  the  exciting 
current  due  to  the  effect  of  the  rotating  component  of  the 
armature's  field  can  be  largely  damped  out  by  suitable  devices. 
Thus  if  the  polar  faces  be  solid,  so  that  the  oscillations  of  the  flux 
produce  eddy  currents  in  the  poles,  the  effect  of  these  currents 
will  be  to  produce  magnetic  fields  tending  to  largely  reduce  the 
oscillations  in  the  flux.  The  same  effect  can  be  more  certainly 
produced  by  suitably  arranged  circuits  placed  on  or  round  the 
poles.  We  will  consider  that  the  reactive  effect  of  these  higher 
frequency  terms  has  been  in  this  way  so  diminished  that  they  are 
negligible. 

It  remains  to  consider  the  fixed  armature  field  of  half 
magnitude  due  to  the  stationary  component  of  the  armature  field. 
In  the  polyphase  machine  the  effect  of  the  armature  when  short- 
circuited  was  due  to  the  two  or  three  circuits,  as  the  case  may 
be,  carrying  the  same  current.  If  the  monophase  machine  be 
equivalent  to  the  polyphase  one,  this  state  of  things  would 
correspond  to'  an  armature  current  in  the  single  circuit  twice  as 
large  as  that  which  circulated  in  each  of  the  circuits  in  the  former 
case.  Under  these  circumstances  the  fixed  component  of  the 
armature  field  would  have  the  same  value  as  in  the  corresponding 
polyphase  case.  Hence  if  we  still  denote  by  Sb  the  back  reactance 
of  the  actual  machine,  and  if  the  ideal  induced  E.M.F.  ^  and  corre- 
sponding short-circuit  currents,  ffig,  be  taken  as  before  (p.  243), 
it  will  evidently  follow  that  the  relation  between  these  and  the 
values  of  the  back  and  armature  leakage  reactances  will  now  be 
given  by 


The  cross  reactance  will,  as  before,  be  given  by 


Hence  the  monophase  machine,  under  certain  conditions,  can 
be  dealt  with  in  the  same  way  as  the  corresponding  polyphase  one, 
and  a  similar  construction  to  that  already  developed  will  hold 
good. 


CHAPTER   XXII. 


ALTERNATORS   IN   PARALLEL. 


Parallel  running  of  alternators.  We  will  now  consider 
the  case  where  two  similar  alternators  are  working  in  parallel  on 
a  pair  of  mains  and  both  delivering  power  to  some  circuit  con- 
nected therewith.  It  is  evident  that  one  condition  that  must 
be  fulfilled  is  that  they  should  be  running  at  the  same  speed, 
and  a  second  condition  is  that  the  state  of  running  should  be 
stable,  that  is  that  any  alteration  from  a  steady  state  of  working 
must  result  in  the  machines  continuing  in  such  a  steady  state. 


We  will  only  take  the  case  where  the  two  machines  are  excited 
so  as  to  produce  equal  nominal  E.M.F.S  and  where  they  are  similar 
in  all  respects,  and  we  will  suppose  that  they  can  take  up 

L.  18 


274  ALTERNATING   CURRENTS 

any  desired  phase  relation  and  yet  continue  to  be  driven  smoothly 
by  the  prime  movers.  In  Fig.  219*  let  OP  and  OPl  be  the 
vectors  representing  the  two  equal  E.M.F.S,  £,  OP  being  that  for 
the  leading  machine,  and  let  the  phase  angle  between  them  be  0. 

As   before  we  will   denote  by  a  the  angle  given  by  tan  a  =  ^  > 

where  R  is  the  armature  resistance  and  8  the  synchronous  reactance 
of  each  armature,  the  impedance  being  /;  the  angle  a  is  in  practice, 
as  we  saw,  somewhere  near  90°.  Let  the  load  be  such  that  the 
angle  of  lag  between  the  current  in  the  outside  circuit  and  its 
terminal  pressure  is  X,  the  resistance  of  the  load  being  RI  and  its 
impedance  Jj.  We  will  first  see  what  relation  must  hold  between 
the  different  quantities  and  then  derive  a  construction  for  the 
same.  Let  the  vector  OA  be  that  for  the  pressure  between  the 
mains,  then  PPl  is  that  giving  the  pressure  tending  to  circulate 
current  locally  round  the  armatures,  AP  is  that  required  to  send 
current  round  the  armature  of  the  leading  machine  and  APl  that 
for  the  lagging  one.  On  these  vectors  draw  two  similar  impedance 
triangles  such  that  the  angles  PAQ  and  P1AQ1  are  each  equal  to  a. 
The  vectors  AQ  and  AQl  can  then  be  taken  to  represent  to 
some  scale  the  two  currents  given  by  the  two  machines  being  R 
times  these  currents,  respectively.  Bisect  Qd  i*1  @i  and  it  is 
evident  that  AGl  will  represent  half  the  current  that  is  flowing 
out  to  the  mains,  or  if  this  be  denoted  by  9$  we  have  AGi  is  \^R. 
Further  the  angle  XAC1}  being  that  between  the  resultant  current 
and  the  pressure,  is  the  angle  X.  Now  bisect  PP±  in  C  and  draw 
the  triangle  A  Gfl.  It  is  clear  that  this  triangle  is  similar  to  either 
of  the  impedance  triangles,  and  that  the  angle  GA  Gl  is  the  angle  a. 
Let  S0  be  the  value  of  the  terminal  pressure  given  by  OA  ; 
since  RI  is  the  resistance  of  the  external  circuit,  we  have 


i         . 
and  hence  we  have  -=  °r 


but  AC  =  AC1/cosa,  which  leads  to 

AC       E.cosX 


O  ~~  2  .  EI  cos  a  ~~  2It  * 

Further  the  angle  CAO,  being  equal  to  (TT  —  CAX)  while  CAX  is 
(GAGj.  -  XAGJ  or  (a  -  X),  will  be  given  by  (TT  +  X  -  a).  It  follows 
that  for  a  given  load  we  can  find  the  position  of  the  point  A  for 
any  assumed  value  of  X  by  means  of  the  following  construction. 

Draw  two  equal  vectors  OP  and  OPl  to  represent  the  two 
equal  E.M.F.S  at  any  phase  angle  6  and  bisect  the  difference  vector 
as  at  C.  Determine  the  point  A  by  making  the  angle  OA  G  equal 

*  Mr  C.  E.  Inglis. 


ALTERNATORS  IN  PARALLEL  275 

CA          I 

to(7r  +  X  —  a)  and  the  ratio  -^-r  equal  to  ^y.     With  a  definite 


condition  of  the  supply  circuit  this  ratio  is  a  constant,  and 
hence  for  any  value  of  the  phase  angle  6  the  point  A  will 
lie  on  the  line  OA,  hence  when  determined  for  one  point,  the 
position  of  A  can  be  found  for  any  other  position  of  the  points 
P,  Pl  by  drawing  a  line  parallel  to  CA  through  the  point  where 
PPl  cuts  OC.  On  the  lines  PA  and  P^A  construct  the  impedance 
triangles  for  the  two  machines  and  project  the  resistance  sides 
on  to  the  pressure  lines  of  the  respective  machines  as  at  MN 
and  MJfi.  These  projections  will  be  proportional  to  the  com- 
ponents of  the  currents  in  the  armatures  that  are  in  phase  with 
these  E.M.F.S.  The  following  expression  can  be  shown  to  give  the 
value  of  the  two  projections  concerned  : 

MN  or 


where  k  denotes  the  ratio  of  AO  to  AC;  a  positive  value  of  ^ 

must  be  taken  for  the  leading  machine,  and  a  negative  one  for 
the  lagging  machine.  Let  the  expression  in  the  bracket  be 
denoted  by  <f>  (0),  then  since  the  power  that  a  machine  is  delivering 
is  equal  to  the  product  of  the  pressure  into  the  in-phase  current 
and  the  latter  is  given  by  MN/R,  while  further  we  have  /  cos  a.  =  R, 
it  follows  that  the  mean  power  will  be  given  by 


The  expression  for  <f>(0)  is  too  complicated  to  be  used  for 
calculation  but  it  can  readily  be  seen  that  it  can  be  written  in 
a  simpler  form.  For  <f>(0)  is  a  function  of  the  sine  and  cosine 
of  half  the  angle  0  and  hence  it  follows  that  it  can  be  written 
in  the  form 


where  the  quantities  £,  77  and  {$  can  be  expressed  in  terms  of  the 
various  constants  of  the  circuits.  The  power  being  thus  repre- 
sentable  by  an  expression  of  this  form  it  follows  that  if  we  can 
determine  three  points  in  the  relation  between  <f>(0)  and  the 
angle  0,  the  whole  relation  can  readily  be  plotted.  The  most 
convenient  points  to  take  are  those  for  which  0  is  zero,  90°  and 
180°,  and  we  will  now  see  how  the  required  curve  can  be  obtained 
from  these  points.  It  is  only  necessary  to  consider  one  of  the 
machines,  since  when  the  curve  for  that  is  obtained  it  will  be  seen 
that  that  of  the  other  readily  follows.  By  carrying  out  the  above 

18—2 


276 


ALTERNATING   CURRENTS 


construction  for  a  special  case  in  which  the  angle  CO  A  was  about 
5°  and  the  ratio  AC/AO  about  9,  the  angle  a  being  60°,  the 
values  of  the  projections  of  the  current  vector  on  the  E.M.F.  vector 
of  the  leading  machine  was  found  graphically  by  the  above  con- 


CA   L 


Fig.  220. 


struction  to  be  given  by  the  lengths  shown  in  Fig.  220,  where  A  B 
is  the  projection  MN  for  zero  phase  angle,  LG  for  90°  and  CD  for 
180°.  The  first  point  is  to  determine  the  value  of  f  in  the 
expression.  We  evidently  have 

AB  =  %  -f  77  sin  /3   and   CD  =  ?  —  77  sin  /£, 
so  that  £  = 


produce  AB  to  E  making  BE  equal  to  CD,  then  it  will  be  seen 
that  AE  =  2f.  Hence  if  AE  be  bisected  in  F  the  line  through 
that  point  is  the  axis  of  the  curve  77  sin  (6  +  0).  The  maximum  of 
this  must  now  be  found.  Since  EF  is  the  value  of  f,  and  BE  is 
f  —  77  sin  j3,  it  follows  that  FB  is  that  of  77  sin  {3.  Further,  the  line 
LG  is  equal  to  the  value  of  f  +  77  sin  ($ -f /3)  when  0  is  90°  or  is 
f  4-  77  cos  /3.  Hence  HL  being  equal  to  f ,  the  part  6r//"  gives  the 
value  of  77  cos  /3.  Thus  if  the  line  HK  is  drawn  perpendicular  to 
GL  and  equal  to  .RF  and  if  GK  is  joined,  it  follows  that  GK  is 
the  value  of  77.  With  a  centre  on  the  line  FK  describe  a  circle 
with  this  radius  and  project  on  to  it  from  the  point  B  marking 
this  point  with  a  0.  Divide  up  this  circle  into  say  twelve  equal 
parts  reckoning  from  0  and  mark  as  shown  with  the  corresponding 
values  of  the  angle,  positive  and  negative.  Then  take  a  base 
line  through  C,  and  in  the  ordinary  way  project  across  to  give  the 
harmonic  curve  determined  by  these  points ;  this  curve  is  shown  to 
the  right.  The  upper  part  will  give  the  relation  between  the 
phase  angle  and  the  power  due  to  the  leading  machine,  the  lower 
that  for  the  lagging  one.  It  will  be  seen  that  the  former  always 
does  more  work  than  the  latter,  and  thus  the  current  that  flows 
between  the  two  tends  to  pull  back  the  leader  and  accelerate  the 
lagger.  Further,  after  a  certain  value  of  the  phase  angle,  the 
lagging  machine  actually  has  negative  power,  this  must  mean 


ALTERNATORS   IN    PARALLEL  277 

that  it  is  receiving  power  from  the  other  in  addition  to  the  line 
power,  or  is  tending  to  act  as  a  motor.  This  action  will  be  referred 
to  at  length  later  on. 

Influence  of  prime  mover.  Since  the  leading  machine 
always  does  more  work  than  the  lagging  one,  it  follows  that  the 
two  machines  tend  to  get  into  such  a  phase  relation  that  the  angle 
6  is  zero.  Under  these  circumstances  the  resultant  vector  PPi 
giving  the  pressure  which  is  causing  current  to  flow  locally  round 
the  two  armatures  is  zero,  and  the  two  machines  are  just  cophased 
on  the  mains  and  antiphased  with  reference  to  one  another.  This 
ideal  state  of  things  cannot  in  general  be  realized,  there  are 
always  slight  differences  in  the  condition  of  the  two  machines 
which  will  tend  at  every  instant  to  disturb  this  state  of 
affairs,  and  hence  there  will  generally  be  a  small  outstanding 
pressure  causing  current  to  flow  between  the  two  armatures ;  this 
current  will  evidently  flow  in  such  a  direction  as  to.  pull  the 
lagging  machine  back  again  towards  the  position  of  zero  phase 
angle  and  may  be  called  the  synchronizing  current  of  the  two 
machines.  Its  value  will  depend  on  the  constancy  of  the  turning 
moments  of  the  two  prime  movers  and  on  the  electrical  constants 
of  the  armatures.  It  is  of  very  great  importance  that  the  prime 
movers  should  exert  as  uniform  a  turning  moment  as  possible  or 
the  synchronizing  current  may  attain  a  large  value  and  the 
regulation  be  badly  affected,  as  well  as  the  maximum  possible  load. 
We  shall  see  when  we  come  to  consider  the  phenomenon  of  hunting 
that  the  matter  is  further  complicated  by  the  fact  that  a  dynamo 
possesses  a  proper  natural  period  of  oscillation  to  and  fro  about  a 
mean  stable  position ;  a  state  of  affairs  in  which  a  machine 
is  oscillating  in  speed  about  a  mean  speed  is  called  "  hunting,"  and 
the  corresponding  change  in  the  current  received  by  it  is  referred 
to  as  "  surging." 

Influence  of  shape  of  curve.  If  the  dynamos  have  different 
shaped  E.M.F.  curves  there  may  be  another  factor  which  will  to 
some  extent  determine  the  value  of  the  circulating  current.  Thus 
let  the  two  machines  have  the  third  harmonic  present  but  in  such 
a  way  that  the  one  is  peaked,  the  other  is  flat  (see  Figs.  90  and  91 ). 
It  will  be  seen  that  in  this  case  although  the  fundamental  sines 
may  be  properly  antiphased  to  one  another,  these  two  harmonics 
will  then  be  cophased  with  regard  to  the  local  circuit  between  the 
armatures  and  will  tend  therefore  to  somewhat  increase  the 
natural  synchronizing  current  that  will  be  necessary  to  keep  the 
machines  in  step.  It  is  found  in  practice  that  even  a  considerable 
want  of  similarity  in  the  E.M.F.  curves  may  not  produce  any  serious 
inconvenience  in  parallel  running. 

Process  of  putting  machines  in  parallel.  We  must  now 
see  how  it  is  possible  to  put  two  alternators  in  parallel  so  as  to 
share  in  supplying  power  to  a  definite  load.  Let  one  of  them,  / 


278 


ALTERNATING   CURRENTS 


(Fig.  221),  be  running,  the  excitation  being  supplied  by  means 
of  the  direct  current  dynamo,  D,  which  may  either  be  a  separate 
little  machine  directly  connected  to  the  field  of  the  alternator 


Fig.  221. 

or,  as  shown,  a  distinct  machine  connected  to  a  pair  of  mains 
or  bus-bars  so  that  it  can  be  used  to  excite  any  one  of  the 
alternators.  To  put  //  in  parallel  with  /  the  first  thing  is  to 
run  it  up  by  its  prime  mover  until  it  is  running  at  what  is 
known  to  be  approximately  the  correct  speed.  The  exciting 
circuit  is  made  and  the  current  adjusted  until  the  voltmeter 
that  is  placed  across  the  armature  shows  about  the  correct 
terminal  pressure,  or  a  little  more  than  that  between  the 
main  bus-bars  B  on  which  the  load  is  placed,  as  shown  by  the 
voltmeter  on  I.  Under  these  circumstances  the  two  machines  / 
and  //  will  be  running  at  nearly  the  same  speed :  let  the  E.M.F.S  be 
the  same  in  magnitude  but  let  the  periods  be  respectively  p 
and  p  +  Sp.  If  //  is  running  faster  than  /,  8p  will  be  positive,  if 
slower  it  will  be  negative. 

It  follows  that  the  difference  between  the  two  pressures,  that 
is  the  pressure  tending  to  circulate  current  between  the  two 
armatures,  will  be  given  by 

E  sin  (p  +  Bp) .  t  —  E  sin  pt, 
that  is  by 


ALTERNATORS  IN  PARALLEL 


279 


Now  let  two  glow  lamps,  each  capable  of  carrying  the  normal 
pressure,  be  placed  as  shown  across  the  switch  blades  of  the 
incoming  machine.  The  above  pressure  will  then  send  a  current 
round  these  lamps  and  this  current  will  be  such  that  its  amplitude 
fluctuates  according  to  the  applied  pressure  difference,  that  is 
according  to  the  expression 


In  other  words  this  current  will  show  the  effect  known  as  "  beats  " 
in  acoustics.  When  the  lamps  glow  brightly  the  two  E.M.F.s  must 
be  adding  as  regards  the  circuit  between  the  armatures,  and  hence 
will  be  opposing  as  regards  the  mains.  On  the  other  hand,  when 
the  lamps  are  black  the  two  machines  are  evidently  antiphased  as 
regards  their  local  circuit  and  hence  cophased  as  regards  the 
mains  ;  this  is  the  condition  that  has  to  be  fulfilled.  Hence  if  the 
incoming  machine,  //,  has  its  speed  carefully  adjusted  till  for  some 
few  seconds  the  lamps  remain  quite  black  we  know  that  the  main 
switch  can  then  be  closed  with  the  machines  at  their  proper  phase 
relation  as  regards  one  another.  The  exciting  current  of  //  can 
then  be  carefully  adjusted  till  it  takes  up  its  proper  share  of  the 
load. 


Fig.  222. 


280 


ALTERNATING   CURRENTS 


This  direct  method  can  be  used  in  cases  where  the  machines 
are  of  low  pressure,  such  as  is  commonly  the  case  in  laboratory 
dynamos,  but  with  machines  of  high  pressure  it  would  be  very 
inconvenient  to  have  lamps  connected  as  shown.  In  such  a  case 
transformers  are  used  as  is  shown  in  Fig.  222 ;  these  transformers 
are  in  addition  often  used  to  operate  the  voltmeters  for  the 
different  machines  instead  of  placing  them  direct  on  the  terminals. 
The  transformers  are  shown  at  T,  T,  and  they  are  connected  up  to  a 
special  pair  of  bus-bars  distinct  from  the  main  pair,  B.  In  this 
case  it  is  unnecessary  to  have  double  pole  switches  for  the 
auxiliary  circuit  and  hence  a  single  lamp  can  be  used  for  each 
machine  as  shown.  In  the  figure  the  exciting  circuits  of  the 
dynamos  are  not  shown.  It  will  be  readily  seen  that  the  machines 
can  be  parallelized  exactly  in  the  same  way  as  just  described  by 
means  of  the  indications  given  by  the  lamps  on  the  secondaries  of 
the  transformers.  In  the  present  case,  however,  it  would  be 
possible  by  joining  up  these  secondaries  in  the  opposite  direction 
for  the  incoming  machine,  to  cause  the  criterion  of  brightness 
rather  than  that  of  darkness  to  be  utilized  as  the  indication  of  the 
proper  phase  relation. 

Many  methods  can  be  used  for  connecting  up  the  parallelizing 
lamps,  especially  in  polyphase  systems.  One  is  shown  in  Fig.  223, 
which  is  practically  identical  with  the  last  described  monophase 


Fig.  223. 

case,  the  lamps  being  merely  fed  by  transformers  placed  across  the 
star  connected  generators.     By  special  connections  it  is  possible  in 


ALTERNATORS  IN  PARALLEL 


281 


the  polyphase  case  to  indicate  the  relative  motion  of  the  two 
machines,  that  is,  to  show  whether  the  incoming  machine  is 
running  too  fast  or  too  slow.  Let  three  lamps  be  connected  to  a 
three-phase  armature  as  shown  in  Fig.  224,  the  usual  transformers 
being  for  simplicity  omitted.  Let  the  vectors  representing  the 


C, 


6 


Fig.  224. 

E.M.F.s  of  the  two  machines  be  at  any  instant  as  shown  in  the 
figure  below,  then  if  the  lamps  be  denoted  as  in  the  figure  by  a,  /9 
and  7,  the  pressure  on  a  will  be  given  by  AQ,  that  on  ft  by  A^C, 
and  that  on  7  by  BBt.  If  the  two  machines  be  running  in 
synchronism  with  the  configuration  shown  in  the  figure,  a.  will  be 
very  bright,  ft  less  bright,  and  7  will  be  very  dim.  When  the  two 
are  in  their  proper  phase  relation,  a  and  (3  will  be  equally  bright 
and  7  will  be  black.  But  if  we  consider  the  vector  system  A^Bfi^ 
to  be  revolving  faster  than  the  system  ABC,  it  is  evident  that  the 
lamps  will  experience  a  cyclic  change  of  terminal  pressure,  and 


282  ALTERNATING   CURRENTS 

will  therefore  brighten  up  in  succession;  further,  if  the  system 
A^Ci  be  rotating  more  slowly  than  ABC  the  same  effect  will  be 
produced,  but  the  lamps  will  now  go  through  their  cycle  of 
brightenings  in  the  opposite  direction.  Hence  the  order  in  which 
they  brighten  will  be  an  indication  of  the  relative  motion  of  the 
two  armatures.  Instead  of  the  three  lamps  it  is  evident  that  a 
suitable  electromagnetic  device  could  be  employed  which  would 
directly  indicate  whether  the  incoming  machine  had  to  be  speeded 
up  or  slowed  when  the  beats  occur. 

In  the  case  of  compounded  alternators  it  is  desirable,  as  in  the 
case  of  direct  current  compound  machines,  to  provide  a  further  set  of 
bus-bars  or  equalizing  bars  connected  with  the  compounding  circuits 
in  order  that  the  excitations  may  not  be  unequally  affected  when 
the  machines  are  put  in  parallel.  These  bars  are  so  arranged  that 
when  the  machines  have  been  put  properly  in  parallel  on  the 
main  bars,  the  series  circuits  are  likewise  put  in  parallel  with  one 
another. 


CHAPTER  XXIII. 

THE  SYNCHRONOUS   MOTOR. 

Action  of  alternator  and  motor.  When  we  were  con- 
sidering the  case  of  two  alternators  working  in  parallel  we  saw 
that  whenever  there  was  a  phase  difference  between  the  two  one 
did  more  work  than  the  other,  and  thus  there  was  a  flow  of  current 
between  the  two  tending  to  bring  the  lagger  into  phase  with  the 
leader.  Now  let  such  a  pair  of  dynamos  be  working  as  before  but 
let  the  external  load  on  the  mains  be  removed.  The  two  will  still 
run  in  parallel.  If  in  addition  the  one  of  them  have  its  prime 
mover  cut  off  it  will  now  be  receiving  current  from  the  other,  and 
in  fact  could  be  loaded  up  mechanically  on  a  brake  and  give  out 
mechanical  power,  all  the  while  continuing  to  run  in  synchronism 
with  the  other.  In  this  case  the  second  machine  is  called  a 
synchronous  motor,  and  we  will  now  proceed  to  investigate  its 
properties.  We  will  for  generality  take  the  two  machines  to  be 
dissimilar  and  to  be  connected  by  a  main  of  definite  impedance : 
they  will  of  course  be  considered  to  be  working  at  the  same 
periodicity. 

For  the  present  it  is  a  matter  of  indifference  whether  the 
machine  be  monophase  or  polyphase,  the  discussion  following  can 
be  taken  to  refer  to  a  single  phase  of  the  latter  machine.  All  the 
phase  relations  per  armature  will  be  the  same  in  a  polyphase 
machine  when  the  loads  on  the  phase  are  balanced,  but  the  power 
absorbed  and  delivered  will  be  greater  in  proportion  to  the  number 
of  phases. 

Vector  relations.  Let  OG  (Fig.  225)  be  the  E.M.F.  of 
the  machine  that  is  acting  as  generator  at  one  end  of  the 
line,  and  let  OM  be  the  E.M.F.  of  the  motor  at  the  other  end 
of  the  same.  These  lines  represent  the  nominal  induced  E.M.F.S 
due  to  the  actual  exciting  currents  as  in  the  case  of  the 
dynamo.  If  the  two  be  running  in  any  stable  manner  the 
two  vectors  will  have  a  definite  phase  difference  that  we  will 
denote  by  6.  What  determines  this  will  be  considered  shortly. 
The  difference  between  these,  that  is,  GM,  must  be  the  pressure 
that  is  necessary  to  send  the  current  down  the  complete  circuit  of 


284 


ALTERNATING    CURRENTS 


the  two  armatures  and  the  line.  If  the  total  resistance  and 
reactance  of  this  is  known,  we  can  draw  on  GM  the  impedance 
triangle  shown  at  GMQ,  such  that  the  side  MQ  represents  the 
pressure  required  by  the  ohmic  resistance,  and  GQ  that  required 
by  the  reactance.  The  three  sides  of  this  triangle  are  severally 
equal  to  <$.R,  <&.S,  and  ^./,  where  ^  is  the  current  flowing, 


Fig.  225. 

R  the  resistance,  S  the  reactance,  and  7  the  impedance  of  the 
whole  circuit.  The  angle  GMQ  is  thus  fixed  when  the  circuit  is 
given,  and  if  we  denote  the  two  E.M.F.S  by  £  and  Ji  it  is  evident 
that  the  three  quantities  <£,  Jl  and  /  remain  fixed  for  all  other 
variations  in  the  relations ;  we  may  call  them  the  characteristic 
quantities  for  the  two  machines  working  on  the  given  line. 

The  direction  of  the  current  is  evidently  that  of  the  vector  00 
parallel  to  MQ  and  in  this  figure  it  leads  both  the  E.M.F  s ;  the 
phase  relation  of  the  current  and  these  E.M.F.s  will  be  considered 
later  on.  It  should  be  noted  that  the  impedance  triangle  for  the 
whole  line  is  really  the  result  of  adding  up  the  several  impedance 
triangles  for  the  different  parts,  thus  if  these  be  GSW  for  the 
generator,  TS  V  for  the  line  and  MTU  for  the  motor,  the  pressure 
at  the  terminals  of  the  generator  will  be  OS  while  that  at  the 
terminals  of  the  motor  will  be  OT.  Hence  the  angles  between 
these  pressures  and  the  current  will  differ  slightly  from  those 
between  the  current  and  the  E.M.F.s,  so  that  a  current  that  is  in 
phase  with  the  generator's  E.M.F.  cannot  be  in  phase  with  the 
pressure  at  its  terminals,  and  hence  the  load  on  the  line  will  be 
inductive.  In  discussing  the  question  it  is  in  general  more 
convenient  to  consider  the  phase  relations  with  reference  to  the 


THE   SYNCHRONOUS  MOTOR 


285 


E.M.F.S,  but   the  above  point  must  be  borne  in  mind  to  avoid 
confusion. 

Expression  for  the  power.  We  can  easily  find  what  power 
is  being  supplied  by  the  generator  and  used  by  the  motor,  for  if 
the  lines  OG  and  OM  be  projected  on  the  direction  of  the  current 
(Fig.  226),  that  is,  on  MQ,  these  projections  are  evidently  such  that 
when  multiplied  into  the  current  they  will  give  the  powers 
required.  Thus  the  power  being  supplied  by  the  generator  will 
be  the  current,  ^,  multiplied  by  PQ,  that  absorbed  by  the  motor 


and  turned  into  mechanical  power  partly  inside  owing  to  friction 
and  core  loss,  and  partly  utilized  outside  in  the  form  of  mechanical 
power,  will  be  given  by  PM  .  <$,  while  that  lost  in  the  resistance  of 
the  whole  line  will  be  given  by  MQ  .  *$.  The  latter  is  necessarily 
given  by  ^R,  where  R  is  the  total  equivalent  resistance  of  the 
circuit  as  before.  From  the  properties  of  the  impedance  triangle 
we  know  that  the  line  GM,  which  we  will  denote  by  &,  has  the 
length  ^  .  /,  and  hence  we  have  ^  =  t£\I.  Thus  the  generator 

.     m      PQ.%  .     m       PM.% 

power  is    Wg  =  ~  j-  —  ,  and  the  motor  power  is  Wm  =  --  j  —  . 

We  can  readily  express  these  quantities  in  terms  of  the 
characteristic  quantities  and  the  phase  angle.  For  draw  the  line 
OH  from  0  making  the  given  angle  a  with  OM,  and  produce  the 
line  MG  to  meet  it  in  H,  calling  the  angle  at  H,  /3.  Since  the 
angles  of  the  triangle  OMH  are  two  right  angles  as  are  those  at 
the  point  M,  it  follows  that  the  angle  OMP  is  also  y3.  Project  the 
sides  of  the  triangle  OMG  on  this  line  and  we  get 


.  cos  a  +  c^.  cos  /S  =  £  .  cos  (a  —  6\ 


286  ALTERNATING   CURRENTS 

but  PM=J£.cosft, 

hence  PM  =  -™  {Jt-£ .  cos  (a  -  0)  -  Jfi .  cos  a}, 

which  leads  to 

Wm  =  r  {^=^cos  (a  -  0)  -  ,//2 .  cos  a}. 


Similarly  we  can  draw  the  line  OK  making  the  same  angle  a  with 
OG,  and  produce  MG  to  cut  it  in  K,  calling  the  angle  at  K,  7. 
By  the  equality  of  the  interior  angles  of  the  triangles  GOK  and 
MGL  we  similarly  see  that  the  angle  at  L  is  also  7.  Hence  if  we 
project  the  sides  of  OMG  on  this  line  we  have 

Ji  cos  (a  +  0)  +  &.  cos  7  =  </.  cos  a,  but  PQ  =  </.  cos  7, 


which  leads  to 

Wg  =  j  (c^2  .  cos  a  -  <J^.  cos  (a  +  &)}. 

Hence  the  powers  have  been  expressed  in  terms  of  the  desired 
quantities. 

It  will  be  noted  that  the  way  in  which  the  powers  depend  on 
the  phase  angle,  a,  is  solely  dependent  on  the  three  characteristic 
quantities,  c^,  Jit  and  a.  The  impedance  only  gives,  so  to  speak, 
the  scale  on  which  the  power  is  to  be  measured,  and  sets  an  upper 
limit  to  its  value.  Thus  the  solution  of  the  question  can  be 
considered  apart  from  any  definite  value  of  the  impedance  of  the 
circuit,  and  will  depend  as  to  its  form  solely  on  the  ratio  of  the 
resistance  and  reactance. 

Example.  Let  us  apply  these  expressions  to  a  definite  case 
lay  way  of  illustration,  taking  for  the  characteristic  quantities 
£=  5000,  Ji  =  4500  and  a  =  60°.  The  results  are  plotted  in  Fig.  227, 
where  the  upper  curve  is  for  the  machine  excited  to  5000  volts 
and  the  lower  for  that  at  4500  volts.  The  ordinates  are  such  that 
they  give  kilowatts  when  multiplied  by  1000/7.  It  will  be  noted 
that  each  has  a  positive  and  a  negative  part.  When  the  upper 
curve  lies  above  the  axis  it  is  acting  as  a  generator,  and  when 
below  as  a  motor,  and  the  reverse  holds  for  the  other  machine. 
There  are  thus  two  regions  of  motor  action,  the  first  when  the 
more  excited  machine  acts  as  generator,  the  other  when  the  less 
excited  one  does  so.  It  will  be  seen  from  the  figure  that  in  the 
former  case  the  motor  action  is  greater  than  in  the  latter  and 
extends  over  a  longer  range  of  phase  angle.  Further,  the  difference 
between  the  ordinates  must  be  the  loss  of  energy  in  the  circuit 
owing  to  resistance,  and  in  the  latter  case  this  is  evidently  greater 
in  proportion  than  in  the  former.  Further,  it  will  be  seen  that 


THE   SYNCHRONOUS  MOTOR 


287 


the  condition  of  the  latter  motor  action  is  associated  with  different 
phase  relations  to  the  current.     These  points  can  be  directly  seen 


30 


Fig.  227. 

as  follows.  Draw  the  triangle  GMQ,  Fig.  228,  as  before,  and  take 
the  case  where  the  generator  has  a  higher  E.M.F.  than  the  motor, 
as  shown  by  the  lines  OG,  OM.  The  power  produced  by  the 
generator  is  proportional  to  PQ  and  that  of  the  motor  to  PM,  the 
loss  being  proportional  to  MQ.  Now  let  the  conditions  be 


P  P'  M        Q 

Fig.  228. 

reversed,  the  higher  excited  machine  being  the  motor,  and  we  get 
the  lines  O'G  and  C/M.  With  the  same  loss,  it  will  be  seen  that 
the  two  powers  are  much  smaller,  being  proportional  to  P'Q  and 
P'M.  Hence  the  condition  of  greater  E.M.F.  in  the  generator  is  in 
general  the  more  efficient  to  employ.  It  will  be  noted  that  in  the 
case  figured,  the  current  leads  on  the  E.M.F.  of  the  generator  when 


288 


ALTERNATING   CURRENTS 


it  is  less  excited  and  lags  on  it  in  the  other  case.  This  lead  or 
lag  is  not  necessarily  associated  with  these  relative  conditions  to 
the  extent  indicated  in  the  figure,  but  a  change  of  phase  relation 
must  ensue.  In  some  cases  the  effect  of  the  over-excited  motor 
tending  to  produce  this  leading  current  may  be  utilized  to  bring 
the  current  and  the  pressure  on  the  terminals  of  the  sending  end 
of  the  line  approximately  into  phase  and  thus  avoid  extra  drop 
in  the  line. 

Stable  action  :  efficiency.  On  referring  again  to  Fig.  227 
it  will  be  seen  that  the  motor  part  of  the  lower  curve  has  two 
equal  parts  which  are  cross  hatched  at  different  angles.  The  first 
part  is  one  in  which  the  motor  can  respond  to  any  required  increase 
in  load  by  properly  drawing  on  the  generator,  and  this  can  be 
done  till  the  maximum  output  of  the  motor  is  attained.  The 
second  part,  however,  indicates  that  with  increase  in  demand  of 
power  from  the  generator,  the  motor's  power  actually  decreases. 
Thus  this  part  of  the  curve  is  unstable,  and  has  no  real  significance 
as  to  the  operation  as  a  motor  capable  of  taking  a  load,  in  fact 


0  0  10  20  30  44  SO  €0  71 

Rel&tire  Angular  Position  of  G  and  M  in  Degrees. 
Fig.  229. 

only  the  first  part  need  be  considered.  Hence  in  Fig.  229  this  first 
part  has  been  drawn  out  to  a  larger  scale  of  phase  angle,  and  in 
addition  the  ratio  of  the  two  powers,  or  the  "  electrical "  efficiency, 
is  shown.  It  will  be  seen  that  the  range  of  useful  phase  difference 
is  again  restricted  by  conditions  of  economy  to  comparatively  few 
degrees,  and  hence  the  actually  useful  part  of  the  curve  is  small. 
The  maximum  efficiency  in  the  case  considered  is  at  about  12  on 
the  power  scale,  or  with  an  impedance  of  20  in  the  line,  is  at 
about  60  kilowatts.  It  will  also  be  noted  that  the  condition  of 
running  light  corresponds  to  a  negative  value  of  the  phase  angle 
of  about  3?°. 


THE   SYNCHRONOUS   MOTOR 


289 


Maximum  output  with  fixed  characteristic  quantities. 

When  the  characteristic  quantities  c^  Jt  and  a  are  given,  certain 
important  points  in  the  working  can  be  found.  The  maximum 
output  of  the  motor  will  evidently  occur  when  cos  (a  —  6)  is  unity 
or  when  (a  —  6)  is  zero  or  TT,  that  is  for  an  angle  of  phase  difference 
equal  to  a  or  to  (a  —  TT).  The  former  corresponds  to  ordinary  motor 
action,  the  latter  to  the  motor  having  the  greater  E.M.F.  In 
the  case  we  have  taken,  the  former  occurs  at  60°,  as  will  be  seen 
from  the  figure.  For  this  value  of  the  phase  angle  the  input  is 


~(£—  .//cos  a),   and   the  input  is      (c^'cosa  — 


2a).     The 


corresponding  values  in  one  case  are  12'5  and  23'75  as  shown  in 
the  figure. 

Zero  output.    The  point  of  zero  output  is  evidently  given  by 
g  cos  (d  —  6)  =  Jl  cos  a, 

or  occurs  at  an  angle  given  by 

.(M        \ 
a  —  cos~M-^  cos  al. 

The  corresponding  input  will  be  found  by  substitution  in  the 
general  expression.  In  our  example  the  phase  angle  for  zero 
output  will  be  found  to  be  60°  -  cos"1  .  0'45  or  -  (3°  .  15'),  which 
agrees  with  the  figure.  The  corresponding  input  is  about  0'2. 


«  M        Q 

Fig.  230. 

The  vector  representation  of  this  case  evidently  corresponds  to 
the  projection  of  OM  on  the  current  vector  being  zero,  and  is  thus 
as  given  in  Fig.  230. 


19 


290  ALTERNATING   CURRENTS 

Electrical  efficiency.     The  efficiency  is  given  by  the  ex- 
pression 

Jt  ^cos  oL  —  6)  —  Jl.  cos  a 


•     ~~~J'  cf.  cos  a  -^  cos  (a  +(9)  ' 
and  hence  the  maximum  efficiency  will  occur  at  an  angle  found 

from  the  relation  -—.  =  0. 
dv 

This  leads  to  the  following  relation  as  giving  the  angle  at 
which  the  maximum  efficiency  occurs  : 

a  +  0)  -  £*  sin  (a  -  0). 


Range  of  motor  action.  The  above  relations  hold  for  the 
case  where  the  three  characteristic  quantities  are  given.  Certain 
other  relations  can  be  found  when  some  of  these  change.  Thus 
we  can  readily  find  the  value  of  the  motor's  E.M.F.  that  will  enable 
it  to  continue  to  act  at  one  as  a  motor.  For  the  motor  will  be 
taking  in  power  all  the  while  the  expression  <^(cos  a  —  6)  —  Jt  cos  a 
is  positive,  and  this  will  have  its  greatest  positive  value  for  a  given 

line  and  generator  E.M.F.  when   the   cosine   is  unity,  hence  the 

G 
greatest  E.M.F.  the  motor  can  have  will  be  given  by  Jt  =  -   —  . 

For  any  excitation  of  the  machine  giving  an  E.M.F.  greater 
than  this  value,  it  will  not  act  as  a  motor. 

Value   of  Jt  for   maximum    power   supplied.     We   can 

similarly  find  the  value  of  the  motor's  E.M.F.  that  will  enable  it 
to  take  in  maximum  power  from  the  generator.  The  maximum 
power  being  proportional  to  Jt\£—  «/^cosa],  this  itself  will  be  a 
maximum  for  variations  of  Jt  when  its  differential  coefficient  with 

G 

respect  to  Jt  is  zero,  this  leads  at  once  to  Ji  =  ~  -   as   the 

value  of  the  E.M.F.  giving  maximum  power.  It  will  be  noticed 
that  it  is  one  half  of  the  maximum  possible  E.M.F. 

It  will  be  readily  seen  that  in  this  case  the  motor's  E.M.F.  is 
equal  to  <$*,  the  length  of  the  line  GM,  which  gives  the  drop  in 


M 

Fig.  231. 

the  line,  and  that  the  generator's  E.M.F.  is   in   phase    with   the 
current.     For  in  any  case   of  maximum  motor  power  we   have 


THE   SYNCHRONOUS   MOTOR  291 

g 

6  =  a  (Fig.  231).    Further  in  the  present  case  we  have  Ji—  ^ . 

But  the  value  of  <•%  is  given  by 

^^  =  jp  +  j*_  2j^,  cos  MOG, 

which  in  this  case  leads  to  Jl  =  &.     Hence  the  angle  OGM  is 
equal  to  a,  from  which  OG  and  MQ  are  parallel. 

Maximum  power  for  given  current.  Another  problem 
easily  solved  is  that  of  finding  the  value  of  Jl  that  for  a  given 
current  will  enable  the  motor  to  take  in  maximum  power.  This 
means  that  the  length  of  the  difference  vector,  $  or  ^ .  /,  is  given, 


Fig.  232. 

and  that  the  projection  of  OM  on  the  current  must  be  a  maximum, 
that  is,  that  E.M.F.  must  be  in  phase  with  the  current.  The  vector 
representation  is  thus  as  in  Fig.  232,  from  which  we  readily  see 
that  the  motor's  E.M.F.  must  be  given  by 

/*  =  Ji*  +  <%*-%£.<%. cos  a. 

It  should  be  noted  that  this  does  not  mean  that  the  current  and 
pressure  in  the  supply  line  are  in  phase. 

Motor  E.M.F.  for  given  output.  Another  important 
relation  is  given  by  the  conditions  that  the  angle  a,  the  generator's 
E.M.F.  and  the  power  taken  from  it  by  the  motor  are  fixed,  and  the 
resulting  relation  between  the  line  current  and  the  motor's  E.M.F. 
is  required.  It  can  readily  be  seen  that  for  a  given  value  of  the 
current,  that  is,  of  ^,  and  also  given  values  of  the  generator's  E.M.F. 
and  the  intake  of  power,  two  values  are  possible  for  the  motor's 
E.M.F.  For  consider  Fig.  233,  where  the  line  MG  has  a  definite 
length,  depending  on  the  current,  and  the  generator  has  a  definite 
E.M.F.  given  by  OG  or  0-fr.  The  constancy  of  the  output  under 
these  conditions  must  entail  the  constancy  of  the  projection  of  the 
motor's  E.M.F.  on  the  line  PMQ.  Hence  if  a  perpendicular  be 
drawn  at  P  the  vector  for  the  motor's  E.M.F.  must  lie  on  this 
line.  If  a  circle  be  drawn  with  centre  G  and  radius  OG  it  will 
cut  this  perpendicular  in  two  points,  0  and  01?  so  that  two  possible 
E.M.F.S  can  exist  for  the  motor,  the  one  OM  smaller  than  OG,  the 
other  OJlf  larger  than  OG.  The  current  in  the  case  shown  leads 

19—2 


292 


ALTERNATING   CURRENTS 


on  the  generator's  E.M.F.  in  the  first  case,  and  lags  on  it  in  the 
second,  and  further  it  is  evident  that  it  will  lag  on  the  motor's 
E.M.F.  while  that  E.M.F.  has  any  value  less  than  the  value  PM,  and 
will  lead  on  it  when  the  E.M.F.  has  a  greater  value  than  PM. 


Fig.  233. 

Again,  the  point  0  corresponds  to  economical  working  as  before 
shown.  We  thus  see  that  for  a  given  power  and  generator  E.M.F. 
there  are  two  possible  motor  E.M.F.S,  the  one  greater  than  the 
generator's  and  associated  with  a  current  leading  on  the  motor 
E.M.F.,  the  other  less  than  that  E.M.F.  and  associated  with  a  current 
lagging  on  the  motor's  E.M.F.  We  must  now  see  how  to  find  the 
relation  required  for  various  intakes  of  the  motor  from  zero  to  the 
maximum  possible  under  the  given  conditions,  namely  that  given 
by  the  relation  g=  1JI  cos  a. 

Construction  for  motor's  E.M.F.*  We  can  show  that  the 
following  construction  will  enable  us  to  find  for  any  given  load  the 
corresponding  values  of  the  motor's  E.M.F.  Ji  and  the  length  ££' 
which  is  proportional  to  the  line  current,  when  the  given  values  of 
cf  and  a  are  known.  Draw  the  line  OG  (Fig.  234)  to  represent  to 
scale  the  E.M.F.  of  the  generator,  and  draw  two  lines  OA  and  AG 
to  meet  in  the  A,  the  angles  formed  with  OG  being  each  a.  Find 
the  value  of  the  given  load  and  let  it  be  denoted  by  k*,  where 
&2  must  be  interpreted  on  the  scale  of  pressure  on  which  OG  is 
measured.  Thus  if  the  power  be  W  watts  and  the  impedance  of 
the  line  be  /,  the  value  of  k2  that  must  be  interpreted  on  the 
pressure  scale  selected  is  W/I.  Take  a  length  equal  to  k  sec  a  and 
draw  a  circle  with  centre  at  A  and  having  the  tangent  from  0  of 
this  amount.  If  any  point  M  be  selected  on  this  circle,  we  shall 

*  Mr  G.  T.  Bennett. 


THE   SYNCHRONOUS  MOTOR 


293 


show  that  the  value  of  the  motor's  E.M.F.  will  be  given  by  the 
length  of  the  side  OM  for  the  given  load,  etc.,  while  the  corre- 
sponding value  of  &,  that  is,  of  /-times  the  current,  will  evidently 


Fig.  234. 

be  GM.     It  will  be  seen  that  for  one  value  of  GM  there  are  two 
possible  ones  for  0 M  as  before  shown. 

Consider  Fig.  235,  where  the  construction  as  far  as  the  triangle 
is  concerned  is  repeated,  and  in  addition  the  former  figure  for  the 
ordinary  relation  between  the  different  quantities  and  their  pro- 
jections is  shown.  We  then  have : 


Fig.  235. 

f  =  PM.MQ  =  OM.  cos  OMP .  MG  cos  a 
=  -OM.cos( OMG + a) . MG  . cos  a 
=  -  OM.  MG  (cos  OMG .  cos  a  -  sin  OMG .  sin  a}  cos  a  ...(1). 


294  ALTERNATING   CURRENTS 

Consider  first  the  expression  OM  .  MG  .  cos  OMG.  From  the 
triangle  OMG  we  have 

OM.MG.  cos  OMG  =  J  (ON2  +  MG*  -  OG*). 
But  since  C  is  the  mid-point  of  the  side  OG,  this  becomes 

(MC*-OC*)  ...........................  (2). 

Again,  from  the  triangle  MA  C  we  have 

MA*  =  AC*  +  MC*  -2AC.  CL, 

where  ML  is  drawn  perpendicular  to  AC.     But  LC  is  equal  to 
MN,  and  thus  we  have 

M  C*  =  MA*  -  AC*  +  2AC  .  MN. 
Hence  (2)  becomes 

MA9  -  AC*  +  2AC.  MN  -  OC*. 
But  AC*  +  OC*  =  OA2,  hence  we  have 

OM.MG.  cos  OMG  =  MA*-OA*  +  2AC  .  MN  ......  (3). 

Now  consider  the  expression  OM.MG.s'mOMG:  it  evidently 
gives  twice  the  area  of  the  triangle  OMG,  and  is  thus  equal  to 
OG.MN;  but  OG  is  twice  GC,  and  hence  it  is  also 

2CG.MN  ...........................  (4). 

Substituting  according  to  (3)  and  (4)  in  (1)  we  have  : 
k*  =  -  {(MA*  -  OA*  +  2AC.  MN)  cos2  a  -  2CG  .  MNsm  a  .  cos  a}. 

But  CG=CAcota. 

Hence 

CG  sin  a  .  cos  a  =  CA  .  cot  a  .  sin  a  .  cos  a  =  CA  .  cos2  a. 
Hence  finally  we  have 


But  since  OA,  k,  and  a  are  constants,  MA  is  constant,  and  thus 
M  describes  a  circle.  From  Fig.  234  it  will  readily  be  seen  that 
(OA*-MA*)  is  (OA*-OT)*  or  is  OT*,  or  if  T  be  the  length  of 
the  tangent  on  the  circle,  k=Tcosa,  or  T=&seca.  Hence  the 
proposition  has  been  proved. 

Relation  between  current  and  motor's  E.M.F.     We  can 

now  derive  the  following  graphical  construction  for  finding  the 
relation  between  the  current  expressed  in  terms  of  £C  and  the 
motor  E.M.F.  Draw  the  triangle  OA  G  (Fig.  236)  as  before. 
Divide  up  A  G  into  a  convenient  number  of  equal  parts  depending 
on  the  different  values  of  the  currents  and  loads  that  are  required 
to  be  considered.  With  A  as  centre  draw  in  a  set  of  load  circles 
numbered  from  /  in  the  figure,  and  with  centre  G  draw  in  a  set 


THE   SYNCHRONOUS   MOTOR 


295 


of  circles  for  the  different  assumed  values  of  $*,  these  circles  are 
numbered  from  1  upwards.  Of  the  power  circles  that  which 
passes  through  the  point  G  will  be  the  one  for  zero  power  and 


130-56  89 

Value  of  Mm  Thousands  of  Vo/ts 


10 


Fig.  237. 


296 


ALTERNATING    CURRENTS 


the  point  A  will  give  the  maximum  power  corresponding  to  the 
given  conditions,  namely  that  for  which  £C  and  Jt  are  equal  (p.  291), 
as  is  evident  from  the  figure.  Fix  on  any  one  of  the  power  circles 
and  mark  the  points  where  it  is  cut  by  the  successive  current 
circles,  the  corresponding  distances  measured  from  0  to  the  points 
of  intersection  will  give  the  values  of  the  E.M.F.  of  the  motor. 
These  can  be  easily  pricked  off  with  dividers  and  transferred  to 
squared  paper  so  as  to  exhibit  the  relation  between  Jt  and  £C  (or 
the  current)  for  each  of  the  power  circles.  A  set  of  curves  thus 
obtained  is  shown  in  Fig.  237  for  the  case  where  the  angle  a  was 
60°  as  before,  the  E.M.F.  of  the  generator  being  5000  volts.  The 
curves  are  a  family  of  quartics  and  they  evidently  consist  of  two 
main  parts,  the  one  convex  to  the  axis  of  pressure  the  other 
concave.  The  upper  part  corresponds  to  the  case  of  unstable 
running  and  has  no  practical  application,  parts  of  the  lower 
portion  of  the  curve  have,  however,  important  properties.  The 
minimum  points  of  these  have  been  joined  by  a  dotted  curve 
which  is  also  a  quartic.  These  points  are  those  for  which  the  line 
current  is  a  minimum  for  the  given  power,  and  at  these  points  the 
current  is  evidently  in  phase  with  the  generator's  E.M.F.  since  at 
them  the  angle  between  the  lines  M G  and  the  current  vector  will 
necessarily  be  the  same  as  that  between  00  and  AG,  namely  a. 
These  dotted  lines  hence  divide  up  the  diagram  into  two  parts,  for 
the  one  the  current  leads  on  the  E.M.F.  of  the  generator,  for  the 
other  it  lags.  The  middle  point  is  that  for  the  maximum  power 
possible  under  the  given  conditions. 


Fig.  238. 


THE   SYNCHRONOUS   MOTOR 


297 


Zero  power  lines.  It  can  readily  be  seen  that  the  quartic 
for  zero  power  consists  in  fact  of  two  ellipses  or  parts  thereof.  For 
consider  the  two  cases  of  zero  power  for  which  the  vector  diagrams 
are  given  in  Fig.  230.  From  the  triangle  GOM  we  have 

c?2  =  Jl*  +  %*  -  ZJl%  cos  GMO. 


But  in  this  case  GMO  is  (90°  +  a)  and  hence  we  have 
JP  ± 


where  the  sign  depends  on  which  figure  is  taken.  Considered  as 
equations  between  ?C  and  Jl  these  denote  two  ellipses  as  shown  in 
Fig.  238.  Since  the  coefficients  of  Jl  and  £C  are  the  same,  the 
axes  of  the  two  lie  at  right  angles  and  at  45°  to  the  axes  of  the 
figure.  It  can  readily  be  seen  that  the  two  axes  of  the  ellipse  are 

and     , 


—  sin  a 


In  the  special  case  for  which  the  complete  set  of  curves  has  been 
drawn  these  become 


5000 


and 


5000 


or   3700   and   13,800, 


Vl'86    '      '   VO'24 
which  can  be  verified  from  the  figure. 

It  is  interesting  to  see  what  form  these  V  curves  take  for 
different  values  of  the  angle  a.  To  show  this,  in  Figs.  239  and 
240  are  given  the  lower  parts  of  two  sets  of  these  curves  for 


7234 

VaJue  ofM  in  Thousands  of  Volts. 


Fig.  239. 

values  of  45°  and  75°  respectively.  It  will  be  noticed  that  the 
curves  get  steeper  the  less  the  angle  a.  Thus  the  more  inductive 
the  circuit  is  the  flatter  are  the  curves. 


298 


ALTERNATING    CURRENTS 


Actual  case.  The  curves  given  in  these  figures  refer  to  an 
ideal  case,  namely  where  the  motor  has  no  applied  load  when  the 
external  load  is  removed,  that  is,  is  devoid  of  internal  loss,  where  it 
can  have  its  E.M.F.  indefinitely  increased  to  any  desired  extent, 


1          3  0-          5  6  8 

Value  of  Mm  Thousands  of  Vo/ts 


Fig.  240. 

where  the  reaction  of  the  machine  is  capable  of  being  represented 
by  a  constant  (the  synchronous  impedance),  and  when  all  the 
quantities  involved  are  sinusoidal.  In  a  real  case  none  of  these 
factors  hold  good.  Even  with  no  applied  load  there  is  an  internal 


18 
16 

I 

\ 

\ 

\ 

\ 

/ 

fCi  ^D 

/ 

/ 

\ 

/ 

oLoa.1 

N$> 

V 

l 

/ 

s 

u 

t-8        1-9         2-0        2-1         22        2-3        2-4        2-5        2-ff 

Amperes  Excitation 
Fig.  241. 

one  due  to  the  friction  and  core  loss,  which  will  vary  with  the 
excitation ;  it  is  not  possible  to  indefinitely  increase  the  E.M.F.  by 
reason  of  the  alteration  in  the  permeability  of  the  magnetic 


THE  SYNCHRONOUS  MOTOR  299 

circuit ;  the  reactance  of  the  armature  is  far  from  a  constant ;  and 
lastly  the  curves  of  pressure  and  current  are  not  sines.  Hence 
the  actual  curves  will  not  be  the  same,  in  particular  the  range 
over  which  they  are  obtainable  is  much  more  restricted  than 
the  figures  show.  It  is  usual  to  plot  these  curves  not  in 
terms  of  the  current  and  motor  E.M.F.  but  in  terms  of  the  line 
current  and  the  exciting  current  of  the  motor.  This  will  be 
dependent  on  the  E.M.F.,  though  of  course  not  linearly,  but  in 
a  manner  which  can  readily  be  found  from  the  saturation  curve 
of  the  machine.  Two  such  curves  for  a  very  small  synchronous 
motor  are  given  in  Fig.  241.  They  show  that  the  general 
character  of  the  relations  are  as  described.  The  well-marked 
minima  fall  to  the  left  as  the  load  is  increased.  In  this  case  the 
unloaded  condition  was  still  one  of  considerably  proportionate 
loading  owing  to  the  rather  large  internal  core  and  friction  load. 
With  large  machines  these  curves  much  more  nearly  approach  the 
ideal  ones.  The  various  points  above  referred  to  cause  the  con- 
ditions of  running  to  be  more  or  less  unstable  long  before  the 
higher  parts  of  the  curve  are  reached. 

Free  periodic  oscillations  of  motor.  In  the  consideration 
of  the  operation  of  the  synchronous  motor  that  has  just  been 
taken  it  was  assumed  throughout  that  the  condition  of  operation 
was  such  that  the  speed  of  the  machine  was  absolutely  constant 
for  each  load.  We  will  now  consider  the  case  where  such  constant 
running  is  in  some  manner  disturbed,  and  for  this  purpose  will 
assume  that  the  nominal  induced  E.M.F.S  of  the  generator  and 
motor  are  both  kept  fixed  in  value,  that  the  power  the  motor  is 
delivering  is  constant  during  any  small  disturbance,  and  that  the 
disturbance  produced  consists  in  a  very  small  alteration  of  the 
phase  angle,  9,  from  the  value  it  must  have  for  the  steady  conditions. 
On  p.  285  it  was  shown  that  for  steady  consumption  of  power  by 
the  motor  the  expression  for  that  power  can  be  written  as  follows, 


T 


.  cos  (a  -  0)  -  JLcos  a}. 


Let  the  equilibrium  of  working  be  suddenly  upset  by  a  small 
alteration  in  the  angle  6  of  the  amount  ?=S#,  the  motor  will 
now  have  a  different  rate  of  working  given  by 


W  +  dW  or    W  +  ^  . 

The  difference  between  these  quantities,  or  f  -^- ,  will  be  power 

which  the  motor  is  either  receiving  in  excess  from  the  generator 
or  is  delivering  thereto,  depending  on  the  sign  of  ?.    This  difference 

can  be  written 

Me 

3in(a-0)  (1). 


300  ALTERNATING   CURRENTS 

In  the  absence  of  any  sources  of  loss  of  energy  this  excess  can 
only  go  to  alter  the  kinetic  energy  of  the  motor's  armature.  Let 
H  be  the  -steady  angular  velocity  of  the  same  just  before  the 
change  and  let  II  (1  -  X)  be  the  value  of  it  at  any  moment  during 
the  subsequent  motion.  The  value  of  X  is  always  small.  Consider 
the  vector  diagram  of  the  E.M.F.S  (p.  284),  which  is  such  that  its 
appropriate  angular  velocity  is  necessarily  p  where  p  is  2?m  ;  the 
angular  velocity,  O,  of  the  armature  will  correspond  to  the  angular 
velocity  p  in  the  diagram,  thus  the  angular  velocity  Xfl  will 
necessarily  correspond  to  one  of  \p  in  the  diagram.  But  this 
must  be  the  rate  at  which  the  assumed  small  alteration,  f,  in  6  is 
changing  after  the  disturbance,  and  hence  we  have 

d?  d\_l   d2Z 

dt~  "-P)  (  '  dt-p'di*' 

The  quantity,  X,  is  a  function  of  the  time,  and  the  kinetic 
energy  possessed  by  the  armature  at  any  moment  will  be  given  by 
^/0O2(1  —X)2,  where  /0  is  the  moment  of  inertia  of  the  armature. 
The  rate  of  change  of  this  will  be 


But  since  \  is  always  small,  if  we  denote  the  normal  kinetic 
energy  in  electrical  units  of  work  possessed  by  the  armature  by 
the  letter  K,  we  have 

*Lir  -      9  v   d^ 

dt  '~dt' 

d   „         2K  d*£ 

which  ogives  -=-  K  =  ----  .  —=—  . 

dt  p     dtf 

But  this  rate  of  change  of  the  kinetic  energy  must  evidently  in 
this  case  be  equal  to  the  above  calculated  difference  of  the 
electrical  rates  of  work,  which  leads  to  the  equation 


, 

(«-*>-^o  ...............  (2). 

Substituting  u*  for      *  ^£_  ™  <«  ~  *>  , 

Z-fl  .  L 

we  finally  get  as  the  differential  equation  connecting  the  phase 
angle  and  the  time  the  following  expression, 


This  equation  has  two  solutions  depending  on  the  sign  of  u\  If 
this  is  negative,  that  is,  if  6  is  >  a,  the  solution  is  exponential  and 
of  the  form  £=  Aeut,  showing  that  the  disturbance  results  in  the 
motor  stopping  ;  in  general  for  stable  working  6  is  <  a  and  thus 
sin  (a  —  6)  is  positive,  the  solution  is  then 

f  =  Z  .  cos  (ut  +  77), 


THE  SYNCHRONOUS  MOTOR  301 

where  Z  is  a  constant,  being  the  amount  of  the  initial  disturbance, 
and  77  is  some  fixed  angle.  This  shows  that  the  disturbance 
results  in  an  oscillatory  motion  with  a  frequency  given  by  the 
coefficient  of  t  in  the  expression  cos  (ut  +  77)  divided  by  2?r.  Hence 
the  frequency  of  the  resulting  oscillatory  motion  is  given  by 


or 


0               ,7         /n.«/&^sin(a  —  0) 
or,  since p  =  ZTTTI,        iv  =  A/  -      far    K   T —  '  '' 

In  the  absence  of  anything  tending  to  damp  out  these  oscillations 
they  would  continue  for  ever  with  constant  amplitude. 

If  instead  of  considering  the  complete  system  formed  by  the 
generator,  mains  and  motor  we  consider  the  restricted  system 
consisting  of  the  latter  only,  which  would  be  possible  in  the  case 
where  the  motors  were  small  compared  with  the  generator,  instead 
of  the  E.M.F.  of  the  generator  we  can  substitute  the  constant 
impressed  pressure,  £Q,  that  is  maintained  at  the  terminals  of  the 
motor,  and  in  that  case  the  value  of  /  is  the  impedance  of  the 
motor's  armature.  Such  a  case  is  afforded  by  the  running  of 
a  rotary  converter  on  constant  pressure  mains.  As  will  be  seen 
in  Chap.  XXIV  the  load  in  that  case  is  the  direct  current  output 
of  the  machine. 

The  above  expression  can  be  written  in  other  approximate 
forms.  Thus  in  general  the  angle  6  is  small  and  the  angle  a  is 
nearly  a  right  angle,  hence  sin  (a  —  0)  is  nearly  unity :  further  the 
E.M.F.  of  the  motor  and  the  applied  pressure  in  such  a  case  will  be 
nearly  equal,  and  if  we  denote  by  ^  the  short-circuit  current  of 

£> 

the  armature  at  full  pressure  it  will  be  given  by  y ,  and  hence 
the  expression  for  the  periodic  time  of  the  oscillations  will  be 

T  = 


If  the  kinetic  energy  stored  be  reckoned  in  joules  and  if  the 
moment  of  inertia  be  denoted  by  M,  we  have  K  =  ^M .  H2.  Or 
putting  $  for  the  revolutions  per  second  made  by  the  armature, 
we  finally  get 


If  the  value  of  M  be  taken  in  kilogram-meter  units  we  must 
multiply  the  value  of  M  in  the  above  by  the  factor  100  and  hence 
we  get  another  form  for  T,  viz. 


302  ALTERNATING   CURRENTS 

This  result  has  been  expressed  in  terms  of  other  quantities 
which  are  in  some  cases  more  easily  applied.  Let  v  be  the  linear 
velocity  of  the  outside  of  the  armature,  d  the  polar  pitch  per  pair 
of  poles,  then  we  can  see  that  the  distance  d  is  traversed  at  the 
velocity  v  in  the  time  1/n  and  hence  v  =  dn.  Again,  let  R  be  the 
radius  of  the  armature  and  let  us  put  m  for  the  quotient  of  the 
moment  of  inertia  M  by  the  square  of  this  radius,  on  substituting 
in  the  above,  noting  that  v  =  2?r .  S .  R,  we  get 


As  an  example  consider  the  case  of  a  three  phase  synchronous 
machine  operating  with  a  mesh  connected  armature  at  a  terminal 
pressure  of  350  volts ;  the  output  is  300  kilowatts,  the  alternations 
25,  the  revolutions  per  minute  500.  The  reactance  of  the  armature 
per  phase  is  0*1  and  the  resistance  0*01,  hence  the  value  of  /  is 
practically  01,  and  the  angle  a.  is  about  84°.  The  moment  of 
inertia  of  the  rotating  part  of  such  a  machine  would  be  about 
8,000  foot-pound  units,  and  hence  the  energy  stored  at  the  given 
speed  would  be 

8000  x  4rc2  x  250,000 

°r   35°'000 


2  x  3600  x  32-2 

Since  a  foot-pound  is  T35  joules,  the  stored  energy,  or  value  of  K, 
is  therefore  475,000  joules.  In  expression  (3),  p.  301,  it  will  be 
seen  that  the  part 


will  be  constant  for  all  conditions  of  operation  and  in  this  case  the 
value  of  it  is 

/3  x  25  x  350 
A/-T—      Atr  rr.-     or   019. 
V    4?r  x  47,500 

The  "3"  is  put  in  as  there  are  three  phases  concerned.    Hence  the 
periods  per  second  of  the  oscillations  will  be  given  by 


019  ^j/l  sin  (a  -  0), 

where  Jl  has  to  be  assumed  and  the  value  of  6  will  depend  on  the 
load  that  is  taken  in  accordance  with  equation  (1). 

Let  the  motor  be  excited  to  350  volts  and  first  take  the  case 
of  light  load,  the  phase  angle  will  then  evidently  be  zero  from 
equation  (1)  and  the  periods  will  be  given  by 

019\/350sin84°    or   019  V350  x  0'9945, 

that  is  3'5.     Let  the  full  load  of  100  kilowatts  per  phase  be  taken, 
then  from  (1)  we  have 

3502  {cos  (a  -  0)  -  cos  a}  =  100,000  x  01, 


THE   SYNCHRONOUS   MOTOR  303 


or  since  a  is  84°  we  have 

cos  («-0)  =  TV<&  +  0104, 


this  leads  to 

cos  (a  -  0)  =  0-185    or   sin  (a  -  6)  =  0*982, 

and  hence  N  =  3*5  nearly,  as  before.  It  will  be  seen  that  the 
value  of  N  is  practically  unaffected  in  this  case,  being  very 
slightly  reduced  in  amount. 

Now  let  the  motor  be  over-excited  and  let  M  be  400  volts. 
At  no  load  we  then  have 

cos  (a  -  0)  =  $$  cos  a  =  0119, 
which  gives  sin  (a  -  0)  =  0*993. 

Hence  N  is  019  V400  x  0*993  =  3*8, 

or  is  somewhat  increased.  As  before  consider  the  full  load  to  be 
taken,  we  then  have 

400  [350  .  cos  (a  -  0)  -  400  (0104)}  =  100,000  x  01, 
which  leads  to 

cos  (a  -0)  =  0190   or   sin  (a  -  0)  =  0*981  ; 


hence  N  =  019  V400  x  0*981  =  3*76. 

The  value  of  N  is  as  before  slightly  diminished  on  loading. 

In  the  case  considered  the  angle  a.  is  nearly  a  right  angle  and 
it  is  seen  that  but  little  alteration  is  produced  in  the  period  by 
means  of  loading.  If  this  angle  is  less  a  more  considerable 
difference  will  be  obtained.  We  will  take  the  same  case  as  the 
last  but  suppose  that  while  the  impedance  remains  the  same,  the 
angle  has  the  value  70°,  the  cosine  of  which  is  0*342.  Such  an 
angle  would  not  occur,  in  all  probability,  in  the  special  case  con- 
sidered. As  before  the  no-load  period  is  given  by 

019  \/350  sin  70°    or   019^329, 
that  is  3*45.     To  find  the  full-load  angle  we  have 

3502  {cos  (a  -  0)  -  0*342}  =  100,000  x  01, 

which  leads  to  cos  (a  —  6)  =  0*423  and  hence  sin  (a  —  0)  =  0*905. 
The  period  is  now  3*2,  showing  a  considerable  diminution. 

Damped  oscillation.  The  oscillations  we  have  just  con- 
sidered would  be  such  as  would  result  from  any  sudden  alteration 
in  the  load  on  the  motor  and  may  be  looked  on  as  similar  to  the 
free  vibrations  of  an  ordinary  mechanical  system.  Hence  if  there 
be  any  opposition  to  the  motion  corresponding  to  ordinary  friction 
in  the  mechanical  case,  the  amplitude  of  the  vibrations  will 
diminish  gradually  in  amount  till  a  new  stable  condition  is 
reached.  It  will  be  readily  seen,  from  the  analogy  with  ordinary 


304  ALTERNATING   CURRENTS 

damped  harmonic  motion,  that  in  order  that  this  effect  may  occur 
it  is  necessary  that  there  be  an  opposing  force  at  each  instant 
which  will  be  proportional  to  the  rate  of  change  of  f,  or  that  the 
differential  equation  must  contain  an  extra  term  and  be  of  the  form 


in  which  case  the  solution  will  be  of  the  form 

?=  e-fl  .Z.cos  {(uz  -frf  t  -  77}, 

showing  that  while  /  has  a  positive  value,  the  amplitude  of  the 
oscillations  will  gradually  diminish.  We  can  readily  see  that  such 
a  positive  term  can  be  produced  in  certain  ways.  In  the  discussion 
of  the  last  case  it  was  assumed  that  on  diminution  of  the  speed 
the  power  demanded  by  the  load  was  unaffected,  that  is  the  power 
taken  was  the  same  when  the  speed  of  the  motor  altered.  In 
general  this  will  not  be  the  case.  If  the  load  is  an  ordinary 
mechanical  one  the  power  demanded  will  be  nearly  proportional 
to  the  speed,  and  if  it  is  electrical,  such  as  is  the  case  with  the 
rotary  converter,  it  will  vary  more  nearly  as  the  square,  since  both 
current  and  pressure  will  increase  with  the  speed.  Hence  if  W 
denote  the  normal  steady  demand  for  power  at  the  normal  speed 
ft  and  if  W  be  the  power  demanded  at  the  increased  speed  ft  +  o>, 
and  further  if  we  assume  the  demand  varies  as  the  rth  power  of 
the  speed,  we  evidently  have 

W-WQ  =  k.(£l  +  a>)r-kMr, 
or  for  small  alterations  of  speed 

W-WQ  =  k.rW-i.co, 
that  is  W-W0  =  C.w. 

d  ft 

But  we  have  ft  is  oc  - 

at 

and  hence  (ft  +  o>)  oc  -^  (d  +  £), 

\AJ\J 

d( 

hence  co  is  oc  -=f  . 

at 

Thus   the  presence  of  a  power  demand  which  is   not   constant 

70 

results  in  the  addition  of  a  term  a  -^  to  our  equation,  where  a  is 

a  constant.  In  the  case  assumed  the  sign  of  the  coefficient 
cr  is  positive  since  increase  of  speed  means  increase  of  work,  and 
thus  a  retarding  effect  on  the  motion.  It  is  possible  to  have 
cases  in  which  this  coefficient  was  negative.  Suppose,  for  example, 
that  the  load  consists  of  a  generator  in  which  the  flux  responds 
slowly  to  the  alteration  of  exciting  current  owing  to  eddy  currents 


THE   SYNCHRONOUS   MOTOR  305 

in  the  field  magnets.  When  the  speed  of  the  motor  falls, 
the  terminal  pressure  of  the  machine  would  fall  and  hence 
the  excitation.  Owing  to  the  eddy  currents  in  the  iron  of  the 
field  magnets,  the  flux,  and  hence  the  E.M.F.,  cannot  fall  to  the 
value  appropriate  to  the  new  speed,  and  hence  it  may  happen 
that  extra  retardation  is  experienced  by  the  motor  when  its 
speed  is  falling  in  this  way  ;  and  conversely,  owing  to  the  delay 
in  the  field  rising  with  increase  of  pressure,  the  motor  would  have 
less  than  its  appropriate  work  to  do  at  any  speed  while  rising  in 
speed.  Thus  the  term  instead  of  having  a  positive  coefficient  has 
a  negative  one.  In  such  a  case  the  solution  of  the  equation  is 


showing  that  the  amplitude  of  the  original  disturbance  goes  on 
increasing  without  limit,  and  hence  eventually  the  condition  of 
working  becomes  unstable. 

The  amortisseur.     There  is  one  method  by  which  a  definite 

j  *j 

positive  value  to  the  coefficient  of  the  -^  term  can  be  produced. 

For  simplicity  take  the  case  of  a  polyphase  machine  in  which, 
as  we  have  seen,  the  armature  currents  tend  to  produce  a  definite 
field  fixed  in  space  relative  to  the  poles  when  the  speed  is 
constant.  If  any  variation  in  the  speed  of  the  machine  occurs 
this  field  will  move  in  space  with  an  angular  velocity  equal  to  the 
change  of  angular  velocity  that  has  taken  place.  It  follows  that 
in  the  present  case  there  will  be  produced  an  angular  velocity  of 

this  field  proportional  to  -g.     This  field  moving  relative  to  the 

poles  will  tend  to  produce  E.M.F.S  in  any  circuits  thereto  fixed,  and 
if  the  poles  were  unlaminated,  would  thereby  produce  eddy 
currents  in  those  poles.  The  consequence  would  be  the  pro- 
duction of  a  torque  due  to  the  reaction  between  the  moving  field 
and  the  currents  produced  thereby  which  torque  would  act  in 
such  a  manner  as  to  oppose  the  change  of  motion.  It  follows  that 
in  such  a  case  there  is  an  expenditure  of  power  which  is  propor- 

7t*  70 

tional  to  the  value  of  ~  or  can  be  written  e  -37, 
at  at 

The  production  of  these  eddy  currents  is  not  desirable  in  the 
poles  themselves,  and  further  owing  to  the  low  value  of  the 
conductivity  such  currents  would  be  comparatively  small  in  value, 
the  form  of  the  paths  in  which  they  flow  would  also  not  be  the 
best  possible  for  the  purpose  of  interreaction  with  the  field  of  the 
armature.  In  modern  machines  the  poles  are  very  usually 
laminated,  but  specially  designed  circuits  are  provided  in  which 
the  desired  currents  can  flow  without  much  loss  of  energy.  This 
is  arranged  by  threading  through  the  poles  sets  of  copper  bars  so 

L.  20 


306 


ALTERNATING   CURRENTS 


as  to  form  a  sort  of  grid  very  like  a  portion  of  a  squirrel  cage 
armature  of  an  induction  motor  as  shown  in  Fig.  242.  In  this  way 
the  eddy  currents  induced  are  constrained  to  flow  in  definitely 
assigned  paths  of  the  best  form  to  produce  the  desired  damping 
effect.  Such  a  grid  is  called  "an  amortisseur."  It  is  evident 
that  there  is  some  best  form  to  give  this  grid,  for  suppose  the 
whole  surface  of  the  poles  to  be  provided  with  a  perfectly  con- 
ducting surface,  then  it  would  be  impossible  for  the  currents 


Fig.  242. 

induced  in  it  to  produce  any  damping  effect  since  no  energy 
would  be  absorbed,  it  follows  that  there  must  be  an  optimum 
arrangement  of  the  amortisseur,  and  this  is  generally  found  by 
experiment.  It  may  be  noted  that  merely  surrounding  the  poles 
with  a  ring  of  copper  will  not  in  general  be  of  much  use,  for  with 
small  limits  for  the  maximum  of  the  oscillations  the  total  change 
of  flux  in  such  a  large  circuit,  due  to  the  angular  oscillation  of  the 
armature,  would  probably  be  very  small,  hence  it  is  necessary  to 
provide  many  possible  circuits  on  the  polar  face  in  order  that  the 
swinging  flux  may  always  find  a  circuit  in  which  to  produce  a 
change  of  flux  and  hence  a  retarding  torque. 

It  may  be  noted  in  passing  that  such  an  amortisseur  circuit 
is  an  additional  preventive  against  damage  should  the  machine 
considered  be  a  dynamo  working  in  parallel  with  others.  For  in 
the  case  of  any  failure  in  the  drive  or  excitation  the  armature  will 
be  supplied  by  the  polyphase  currents  and  will  act  in  the  same 
way  as  the  stator  of  an  ordinary  induction  motor,  the  bars  on  the 
field  magnets  forming  a  squirrel  cage  rotor,  and  hence  the  machine 
will  run  on  as  an  induction  motor  without  any  danger. 

The  equation  (p.  304)  now  takes  the  form 


where  a  and  e  are  the  coefficients  just  found.     This  reduces  to 


THE  SYNCHRONOUS   MOTOR  307 

or  writing  f  =  ^-~  -  ',  it  reduces  to  the  form  given,  and  hence 

has  the  solution  there  indicated. 

The  frequency  is  thus  slightly  different  from  the  case  where 
damping  is  absent,  but  the  quantity  /is  very  small  compared  with 
u  in  any  practical  case,  and  hence  the  new  frequency  is  practically 
the  same  as  that  in  the  previous  case. 

It  will  be  noticed  that  e  increases  with  the  load,  for  the  field 
due  to  the  armature  increases  with  the  current  and  with  that 
field  will  increase  the  currents  induced  in  the  amortisseur,  hence 
we  may  approximately  say  that  the  value  of  e  increases  as  the 
square  of  the  load,  and  thus  stability,  depending  on  the  amount  of 
the  damping,  will  increase  greatly  as  the  machines  are  loaded  up. 

Forced  oscillations.  In  addition  to  these  free  oscillations 
of  the  armature  we  may  have  others  corresponding  to  the  forced 
ones  of  a  mechanical  system.  Such  periodic  impressed  forces  can 
arise  in  many  ways,  such  as  from  a  varying  turning  moment  of  the 
prime  movers,  the  hunting  of  the  governors  of  the  same  etc.  The 
effect  will  be  to  produce  forced  oscillations  of  the  armature  and 
the  conditions  can  be  found  as  follows. 

Let  the  power  supplied  to  the  motor  have  a  periodic  term 
superposed  on  the  necessary  constant  term  that  corresponds  to  the 
constant  load  on  the  same,  and  let  this  periodic  power  be  given  by 
Q  .  sin  qt.  Then  instead  of  equating  the  excess  power  of  the 
motor  to  the  increase  in  kinetic  energy  of  the  same  as  in 
equation  (2),  we  must  equate  these  terms  and  this  periodic  one  ; 
this  leads  to  the  equation 


for  the  determination  of  the  resulting  motion,  where  for  the  sake 
of  shortness  the  letter  P  is  used  for  the  expression  in  equation  (1). 
The  letter  e  denotes  the  value  of  the  coefficient  concerned  in  the 
dissipation  of  energy  in  the  polar  faces  as  just  described,  which  we 
saw  was  proportional  to  the  rate  of  change  of  the  latter  angle  f. 
The  solution  of  a  differential  equation  of  the  form 


Hence  in  this  case  we  have  the  solution  in  the  form 

sin  (qt  —  77), 


20—2 


308  ALTERNATING   CURRENTS 

and  in  addition  there  is  necessarily  the  "complementary  function" 
which  in  this  case  is  merely  the  equation  for  the  free  oscillations 
which  we  will  consider  as  damped  out.  The  phase  angle,  ij,  has 
no  special  interest  for  the  purpose  of  the  problem.  The  period  of 
the  impressed  motion  is  thus  the  same  as  that  of  the  fluctuation 
and  its  amplitude  is 


This    can    be   written   in   a   more   convenient   way.       For   from 
equation  (3)  it  will  be  seen  that 


P.  2K 


where  N  is  the  natural  period  of  the  motion  assumed  undamped,. 
since,  as  before  mentioned,  with  the  moderate  damping  that 
occurs  in  these  cases  the  period  of  the  damped  oscillations  will  be 
very  nearly  the  same.  Again,  if  M  denote  the  period  of  the 
impressed  fluctuation  we  have  q  =  2?rif  ,  and  thus  the  expression 
for  the  amplitude  becomes 

Q 


Consider  the  effect  of  varying  the  moment  of  inertia  of  the  motor. 
When  this  is  very  small  its  natural  period  is  very  high,  and  hence 

the  term  -~  will  be  negligible  and  the  amplitude  will  be  given  by 


Now  let  the  moment  of  inertia  be  gradually  increased;  the  quantity 
(1  —  -T^J  will  diminish,  and  hence  the  amplitude  increase  till, 
when  the  two  periods  M  and  N  are  equal,  the  amplitude  is  given 
by  —  or  is  only  restrained  by  the  eddy  current  action  from  being 

infinite.  In  general  the  amplitude  due  to  such  a  condition  would 
be  so  large  as  to  prevent  the  present  theory  holding,  and  the 
oscillations  would  be  so  great  that  the  machine  would  fall  out  of 
step,  being  carried  outside  the  possible  range  of  stability,  hence  this 
resonance  between  the  natural  and  forced  periods  must  be  carefully 
avoided.  With  still  further  increase  in  moment  of  inertia  the 
amplitude  will  go  on  decreasing  continuously.  In  most  ordinary 
cases  the  natural  period  of  the  machine  falls  well  outside  the  latter 
limit  as  compared  with  that  of  any  periodic  variation  in  the  turning 


THE   SYNCHRONOUS   MOTOR  309 

moment  of  the  prime  movers  driving  the  generators.  In  the 
event  of  any  approach  to  a  condition  of  resonance  the  natural 
period  of  the  motor  can  be  altered  by  means  of  either  increased 
fly-wheel  effect  or  by  altering  the  value  of  the  quantity  P  in  the 
equation  for  the  amplitude.  This  may  be  done,  as  will  be  seen  by 
reference  to  equation  (3),  by  either  altering  the  excitation  or  by 
altering  the  value  of  the  impedance  in  circuit  between  the 
machines  by  the  insertion  of  reactance. 

The  condition  of  good  running  in  a  synchronous  machine  thus 
resolves  itself  into  two  parts.  To  ensure  stability  as  regards  the 
free  oscillations  efficient  damping  must  be  provided  by  the  use  of 
appropriate  circuits  on  the  poles,  while  to  ensure  absence  of 
trouble  from  the  forced  oscillations,  care  must  be  taken  to  so 
arrange  the  natural  period  of  the  machine  that  it  is  far  from  being 
near  any  of  the  possible  periods  that  may  arise  in  the  prime 
movers. 


CHAPTER  XXIV. 

THE   ROTARY  CONVERTER. 

A  VERY  important  form  of  synchronous  motor  is  that  known  as 
the  Rotary  Converter.  Let  us  suppose  that  we  provide  an 
ordinary  direct  current  dynamo  with  two  slip  rings  attached  to 
two  opposite  points  of  the  armature.  Then  brushes  attached  to 
these  rings  will  deliver  an  alternating  current  the  periodicity  of 
which  will  be  the  same  as  the  number  of  rotations  per  second 
made  by  the  dynamo.  Hence  we  could  use  such  a  machine  to 
transform  direct  currents  into  alternating  by  merely  driving  it 
from  a  direct  current  source  of  energy;  the  periodicity  of  the 
current  would  depend  on  the  speed  of  the  machine  and  could  be 
adjusted  by  altering  the  exciting  current  by  means  of  the  usual 
shunt  regulating  resistance,  the  applied  direct  pressure  being 
constant.  On  the  other  hand  we  may  supply  alternating  currents 
to  the  slip  rings  and  let  it  run  as  a  synchronous  motor,  care  being 
taken  to  get  it  in  the  proper  phase  relation  in  the  way  already 
described,  the  speed  of  the  machine  being  regulated  by  the  shunt 
resistance.  Under  these  circumstances  we  could  take  direct 
currents  out  of  the  ordinary  commutator,  and  thus  turn  alternating 
currents  into  direct.  In  the  latter  case  it  is  evident  that  the 
speed  must  remain  constant  being  the  synchronous  one;  no 
alteration  of  speed  will  be  produced  by  adjustment  of  the 
excitation  by  means  of  the  shunt  resistance,  but  from  what  we 
have  previously  seen,  such  adjustment  will  result  in  alterations  of 
the  phase  angle  between  the  alternating  current  and  pressure. 

E.M.F.  relations.  Several  points  must  be  considered,  thus 
in  the  case  we  have  taken  of  a  machine  with  two  opposite  slip 
rings  it  is  evident  that  the  maximum  of  the  alternating  E.M.F.  is 
equal  to  the  value  of  the  applied  direct  current  one,  hence  the 
virtual  alternating  pressure  will  necessarily  be  less.  The  ratio 
that  the  virtual  alternating  pressure  bears  to  the  direct  one 
depends  partly  on  the  form  of  the  induction  curve  of  the  pole 
faces.  The  simplest  law  to  assume  is  that  the  induction  through 
any  plane  in  the  armature  passing  through  the  shaft  is  a  sine 
function  of  the  time.  This  is  not  accurate,  and  various  laws 


THE   ROTARY   CONVERTER 


311 


connecting  the  angle  and  the  flux  can  be  obtained  by  altering  to 
some  extent  the  angle  of  the  pole-pieces.  On  this  assumption  of 
a  sine  distribution  we  see  that  the  alternate  pressure  for  the  case 
of  two  slip  rings  will  be  l/\/2  times  the  direct  current  one. 

If  we  place  three  such  rings  attached  to  points  equally  spaced 
round  the  armature,  three-phase  currents  will  be  obtained  from 
them.  Again,  if  four  rings  be  used,  at  45°  pitch,  we  can  use 
opposite  pairs  to  deliver  two-phase  currents. 

In  Chap.  IX  we  considered  the  value  of  the  E.M.F.s  produced  by 
a  distributed  winding  in  an  alternating  current  machine,  but  it  is 
more  convenient  to  again  derive  the  relations  at  this  point. .  Take 
as  an  example  a  two-pole  ring  armature  as  shown  in  Fig.  243,  and 
let  it  be  assumed  that  the  flux  from  the  poles  into  the  armature 
is  such  that  the  amount  entering  any  single  turn  is  a  sine  function 
of  the  angle  which  that  turn  makes  with  the  axial  line  OA,  that 
is  if  a  turn  is  at  the  point  Q  the  flux  in  it  will  be  proportional  to 
the  sine  of  the  angle  between  the  line  OA  and  the  line  OQ  or 


Fig.  243. 

proportional  to  sin  <£.  It  follows  that  the  E.M.F.  induced  in  that 
turn  when  it  is  rotating  with  uniform  velocity  will  be  proportional 
to  cos  c/>,  or  can  be  written  as  e .  cos  (/>,  where  e  is  the  maximum 
E.M.F.  in  a  turn,  or  that  produced  when  the  turn  is  on  the  line  OA. 
Consider  a  small  coil  subtending  the  angle  d<f>,  and  let  the  turns 
per  radian  be  er.  The  E.M.F.  in  the  elementary  coil  at  Q  will  then 
be  ae .  cos  (f> .  d(f>  for  e  cos  c/>  is  the  E.M.F.  in  one  turn  and  a .  d<j>  is 
the  number  of  turns  in  the  coil.  If  it  is  required  to  find  the 
direct  current  E.M.F.  this  expression  must  be  integrated  between 

the  limits  -=  and  —  ^  which  leads  to 


E 


f5 

=  o- .e  \ 

J   w 


COS  (f)  .  d<f)  =  2(7  .  6. 


This  expression  agrees  with  the  usual  formula  for  the  E.M.F.  of 
a   direct   current   dynamo.     Consider   still   the   case   where   the 


312  ALTERNATING   CURRENTS 

armature  is  of  the  ring  form,  and  let  the  maximum  flux  passing 
across  the  armature  be  O.  The  maximum  flux  through  any  coil 
will  then  be  O/2.  Let  the  armature  make  n  revolutions  per 
second,  then  since  the  flux  is  assumed  distributed  in  a  sinusoidal 
manner  the  value  of  the  maximum  E.M.F.  in  a  coil  is  given  by 

<l> 
e  —  27rn  .  -=-  or  7rn<&.     Hence  we  have  E  =  2?r  .  n  .  a  .  <E>.     But  if  z 

denote  as  usual  the  number  of  peripheral  conductors  we  have 
z  =  27Tcr,  and  hence  E  =  <&nz,  which  is  the  usual  form  of  the 
equation  for  the  E.M.F.  of  a  direct  current  machine. 

Now  let  two  points  be  taken  on  the  winding  such  that  the 

angle  between  them  is  -  -  where  m  is  an  integer,  to  each  such 

point  can  be  attached  a  ring  on  which  presses  a  brush  as  before 
considered,  and  there  will  be  m  rings  in  all.  In  such  a  coil  will  be 
produced  an  alternating  E.M.F.,  and  it  is  evident  that  this  E.M.F. 
will  have  its  maximum  when  the  axis  of  the  coil,  OM,  lies  on 
the  line  OA.  The  value  of  this  maximum  E.M.F.  is  evidently 
given  by 


rm 
.e  I 

J 


cos  .  <f>  . 


.        7T 

or  is  Em  =  zo-e  .  sin  —  . 

m 

If  the  coil  has  its  axis  at  the  angle  0  to  OA  as  shown  by  OMl  the 
E.M.F.  will  then  be 

Em  cos  6  or  Em  cos  .pt, 

and  hence  the  E.M.F.  will  be  a  simple  harmonic  one  with  a  period 
equal  to  the  turns  per  second  made  by  the  armature. 
The  virtual  value  Sm  of  this  E.M.F.  will  be 


or  if  we  express  it  in  terms  of  the  direct  current  E.M.F.,  E,  pro- 
duced by  the  same  machine  we  have 

E       .       7T 
0m  =  -rb>  SID.  -  . 

\/z        m 

From  this  we  readily  deduce  the  following  numbers  for  the  virtual 
pressure  existing  between  adjacent  rings  in  such  an  armature. 

No.  of  rings  m  E.M.F.  ratio 

2  ^  =  0707 

3  J  ,/f  =  0-612 

4  J=0-5 


THE    ROTARY   CONVERTER 


313 


In  practice  the  rings  are  some  multiple  of  2  or  3  so  as  to 
permit  the  machine  being  used  on  ordinary  polyphase  systems. 

A  simple  two-pole  rotary  would  have  to  run  at  far  too  high  a 
speed  when  it  is  of  other  than  a  small  size,  and  hence  such  machines 
are  almost  universally  multipolar  ones. 

Vector  representations  of  E.M.F.s.  The  relations  between 
the  values  and  phases  of  the  E.M.F.S  in  a  rotary  converter  are  well 
seen  by  reference  to  the  corresponding  vector  representations. 
For  example  let  a  circle  be  taken,  Fig.  244,  whose  diameter  A  B 
is  equal  to  the  direct  current  E.M.F.  E,  and  let  an  equilateral 
triangle  be  drawn  in  it,  then  the  maximum  of  the  E.M.F.S  between 
the  rings  attached  to  the  three-phase  converter  will  evidently 
be  given  by  the  lengths  of  the  sides  of  this  triangle,  and  since 

the    virtual  E.M.F.  is   -^  times  the  maximum  on  our  sinusoidal 

V^ 

assumption,  it  follows   that   the  E.M.F.  between  adjacent  points 


in  the  three-phase  case  will  have  a  maximum  value 


fi1  /*-{ 


and 


a  virtual   value 


0    .    . 

—  \  — 


In   the   same   way   Fig.   245   gives  the 


vector  diagram  of  the  four-phase  case,  and  it  is  very  simple 
to  verify  the  relationship  between  AC  and  AB  that  we  have 
just  obtained  for  this  case.  Similarly  Fig.  246  shows  the  six- 
phase  case,  from  which  the  various  possible  connections  can  be 


seen  at  a  glance.  In  all  these  cases  the  student  will  notice  that 
the  centre  of  the  circle  is  a  point  of  symmetry,  in  the  same  way 
that  the  various  connections  of  polyphase  transformers  considered 
in  Chap.  XIII  had  a  neutral  point  or  point  of  symmetry.  On  com- 
paring the  diagram  of  the  potentials  given  in  that  chapter  it  will 
readily  be  seen  that  they  are  such  in  relative  magnitude  and 
phase  as  to  render  them  suitable  for  connecting  to  machines  of 
the  type  we  are  considering,  and  in  fact  this  is  the  chief  use  of 
many  of  the  arrangements  therein  considered.  The  fact  that  both 
the  transformers  and  the  machines  possess  definite  neutral  points 


314  ALTERNATING   CURRENTS 

is  one  that  it  is  important  to  keep  in  mind.  Suppose  that  by 
some  want  of  symmetry  the  two  points  are  not  always  at  the  same 
pressure,  then  it  is  evident  that  this  varying  pressure  between  the 
two  neutral  points  will  cause  currents  to  flow  in  the  system 
in  addition  to  those  incidental  to  the  working  of  the  machines. 
It  is  of  course  rare  for  there  to  be  any  great  want  of  symmetry  in 
such  cases,  but  exigencies  of  manufacture  and  other  circumstances 
may  cause  small  deviations  from  the  ideal  symmetry  that  we  have 
considered  (see  Chap.  XI)  and  lead  to  the  existence  of  these 
parasitic  currents.  As  another  example  of  their  possible  occur- 
rence take  the  case  of  a  converter  working  from  a  dynamo  of  higher 
pressure  by  means  of  auto-transformers.  In  this  case  it  will 
evidently  be  necessary  that  the  neutral  points  coincide  throughout 
the  system,  hence  the  leads  from  the  auto-  transformer  must  be 
taken  off  from  it  symmetrically  with  reference  to  its  middle  point 
in  order  to  avoid  the  flow  of  the  balancing  currents.  If  the 
connection  between  the  two  be  by  means  of  tranformers,  no  such 
difficulty  is  experienced,  since  it  is  then  impossible  for  the  parasitic 
currents  to  flow,  all  that  can  happen  is  a  variation  of  the  potential 
of  one  or  other  neutral  points. 

Current  relations.  The  question  of  the  magnitude  of  the 
currents  that  will  flow  in  the  different  sections  of  the  armature 
must  now  be  considered.  As  an  approximation  first  take  the 
case  where  the  load  carried  on  the  alternate  current  side  is  non- 
inductive  so  that  the  power  factor  is  unity  when  the  load  is,  as  it 
must  be  in  a  symmetrical  converter,  a  balanced  load.  If  C  denote 
the  direct  current  flowing  under  the  E.M.F.  E,  and  if  ^m  be  the 
virtual  value  of  the  alternate  current  in  any  one  of  the  sections  of 
the  armature  flowing  under  the  virtual  pressure  Sm  ,  and  if  farther 
we  neglect  the  ohmic  losses  in  the  armature,  we  must  evidently 
have  EG=  m  .  £n$)m,  where  m  is  the  number  of  tapping  points  on 
the  armature.  Hence  we  can  approximately  write 


-         or       ,- 

m  S  m  m 


In  the  ordinary  cases  it  is  easy  to  find  the  line  currents  in  the 
same  way  as  the  corresponding  cases  of  the  ordinary  three-phase 
circuit.  If  the  power  factor  is  not  unity  it  is  evident  that  the 
currents  will  be  increased  in  the  proportion  of  the  secant  of  the 
angle  of  phase  difference. 

Reaction  and  ohmic  loss.  Two  points  must  now  be 
considered.  Firstly,  it  is  evident  that  the  armature  reactions 
of  the  direct  and  alternating  currents  oppose,  since  the  one  set 
act  as  generator  currents,  the  other  as  motor  currents,  also 
the  actual  current  in  any  wire  is  the  difference  of  the  two 
sets  of  currents,  consequently  we  may  make  the  armature  far 


THE   ROTARY   CONVERTER 


315 


stronger  magnetically  than  we  could  with  an  ordinary  direct 
current  machine  of  the  same  output,  though  the  commutator 
must  be  just  as  large.  The  large  number  of  conductors  on 
the  armature  will  enable  us  to  use  a  very  much  smaller  field 
magnet,  since  the  flux  of  magnetism  can  be  reduced  for  the 
same  E.M.F.  In  fact  a  very  important  factor  in  the  design  of  a 
rotary  converter  is  the  peripheral  speed  allowable,  rather  than  the 
conditions  of  current  collection.  The  second  point  is  also  con- 
nected with  the  differential  action  of  the  currents,  and  is  the 


question  of  ohmic  loss.  We  will  take  the  case  of  unity  power 
factor  and  will  investigate  the  relative  ohmic  losses  in  the 
armature  when  it  is  used  as  a  rotary  converter  with  in  rings,  and 
when  used  as  an  ordinary  direct  current  generator.  Let  Fig.  247 
represent  one  coil  of  the  armature  and  let  P  and  P^  be  the  two 
end  wires  in  one  of  the  m  sections  of  the  winding,  if  there  be 
m  rings  the  angle  POPl  will  be  27r/m  for  a  two-pole  machine. 
Let  OM  be  the  central  line  of  the  section  and  consider  the  current 
in  a  wire  at  the  point  Q  which  is  at  such  a  position  that  QOM  is 
a  definite  angle  <£;  further  at  the  instant  considered  let  the 
central  section,  OM,  make  the  angle  6  with  the  axis,  AO,  of  the 
machine.  In  the  wire  at  Q  there  will  coexist  two  currents,  a 
direct  one  and  an  alternating  one ;  if  the  direct  current  that  the 
armature  is  producing  be  denoted  by  C  that  in  the  wire  will 
be  (7/2,  and  if  ^  be  the  virtual  value  of  the  alternating  current  in 
the  wire  its  maximum  will  be  \/2  .  ^. 

For  the  sake  of  completeness  let  us  suppose  that  the  circuit  is 
balanced,  but  that  there  is  a  definite  phase  angle,  A,  then  from  p.  314 
it  will  be  seen  that  the  maximum  value  of  the  alternating  current 

will  be  —  cosec  —  .  sec  X .  C  where  C  is  the  direct  current  flowing- 
m  m 


316  ALTERNATING   CURRENTS 

up  to  the  brushes,  BJ5.  The  current  in  any  one  of  the  wires  of 
the  armature  can  then  evidently  be  written 

c  =  •=•  \  1 cosec  —  sec  X  cos  (6  —  X) 

2  (        m  m 

since  half  the  direct  current  flows  down  each  half  while  the 
alternate  current  is  that  actually  flowing  in  the  section  PP^, 
For  the  sake  of  shortness  write  this  in  the  form 

c  =  ^{l-a.cos(0-X)}. 

In  order  to  find  the  mean  rate  of  production  of  heat  in  the  coil 
PPi  we  must  'first  find  the  mean  square  of  the  current  for  the 
single  wire  at  Q  as  the  coil  turns  through  half  its  revolution,  and 
then  find  the  mean  value  of  this  for  the  different  wires  con- 
stituting the  coil. 

To  find  the  first  mean  value  we  must  integrate  c2  over  the 
angle  from  the  point  where  M  coincides  with  brush  Bl  to  the 
point  where  it  coincides  with  brush  B  and  divide  by  the  angle 

traversed.     The  two  limits  are  evidently  i  -=  —  cf>  J  and  —  i  —  +  <p ) 

while  the  angle  traversed  is  TT.  Hence  the  mean  value  of  the 
square  of  the  current  in  the  wire  0  as  it  rotates  will  be 

L .  91  [  2         {i  _  2a .  cos  (0  -  X)  +  a2 .  cos2  (0  -  X)}  dd. 

-O) 

The  indefinite  integral  is 


and  hence  on  substitution  it  will  be  seen  that  the  mean  value 
reduces  to 


7T  '         2 

We  must  now  find  again  the  mean  value  of  this  for  the  whole 
coil  PPl.     To  do  this  the   expression  must  be  integrated  over 

77"  77" 

the  angle   6  between  the  limits  H  —  and   --  and  divided  by 
2-7T  mm 

—  .     It  is  evidently  only  necessary  to  consider  the  part  con- 

taining  the   angle    <£   and   the    value   of   the   integral   for   that 
part  is 

7T 

>  \  7  ,       m  f  .     /7T         \        .     /TT 

X)  d*  =  -   sm    ~  +  X   +  sm    -  -  X 


m    .    IT 

=  —  sin  —  cos  X. 
TT        m 


THE    ROTARY   CONVERTER  317 

Hence  the  final  mean  value  for  the  whole  coil  reduces  to 

a2) 


or  substituting  for  a  we  finally  get 

C2  L      16       8  .  TT 

-j-  \  1 -I — -  cosec2  —  sec2 

4  [        7T2      m2  m 

as  giving  the  mean  rate  of  generation  in  the  coil.  The  heat 
generated  in  a  simple  direct  current  armature  would  be  simply 

(/^\  2 
-~  J ,  hence  for  the  same  heating  limits  it  is  evident 

that  the  reciprocal  of  the  expression  in  the  brackets  will  give  the 
relative  loads  that  can  be  carried  by  converters  with  different 
number  of  rings,  in  terms  of  the  corresponding  direct  current  load. 
The  case  where  the  load  is  practically  non-inductive  is  the  most 
interesting;  in  this  case  the  ratios  are  as  given  below. 

Direct  current         2  rings        3  rings        4  rings         6  rings 
1-00  0-85  1-32  1-62  1'92 

Since  the  ratio  depends  on  the  secant  of  the  angle  of  lag  it 
will  be  seen  that  a  leading  or  lagging  current  soon  brings  down 
the  possible  load  of  the  machine.  Thus  to  find  the  angle  of  lag 
for  which  a  three-phase  rotary  will  have  the  same  current  carrying 
capacity  as  the  corresponding  direct  current  machine  we  have  the 
equation 

-f  =  ?  cosec2. 60°. sec2  X, 
TT^     y 

which  leads  to  cosX  =  0'8   or  X  =  36°. 

It  will  be  noticed  that  the  capacity  increases  rapidly  with  the 
rings  and  for  example  a  6-ring  rotary  has  nearly  double  the  output 
of  the  corresponding  direct  current  machine.  The  use  of  the  three 
to  six-phase  transformations  on  p.  160  will  now  be  seen.  Which 
of  the  different  methods  there  described  for  obtaining  six-phase 
from  three  is  used  is  a  matter  of  convenience.  The  diametral 
has  the  advantage  over  the  double  delta  in  that  it  enables  one 
to  use  higher  pressures  and  thus  reduces  the  size  of  the  leads 
necessary. 

Starting.  We  must  now  briefly  consider  the  question  of 
starting  up  a  rotary.  Let  the  rings  of  such  a  machine  be  con- 
nected to  the  mains,  the  alternating  currents  in  the  armature  will 
induce  currents  in  the  pole-pieces  of  the  machine,  and  there  will 
be  a  torque  produced  just  as  in  the  induction  motor,  but  the  pole- 
pieces  are  unfavourably  formed  for  such  currents  to  be  very 
effective  in  producing  a  torque,  so  that  although  the  torque  is 
sufficient  to  speed  up  the  machine  to  synchronism,  when  the  field 


318  ALTERNATING   CURRENTS 

circuit  can  be  closed  and  the  machine  will  act  as  a  direct  current 
one,  it  is  at  the  expense  of  a  very  great  call  on  the  mains  for 
current.  This  causes  bad  regulation  in  the  supply  system.  It  is 
possible  to  largely  diminish  this  current  by  means  of  the  starting 
auto- transformers,  one  in  each  phase,  that  have  been  already 
described  in  connection  with  the  induction  motor.  But  there  are 
two  other  difficulties  in  this  method  of  proceeding.  Firstly  when 
the  machine  is  started  in  this  manner,  the  field  circuits  being  open 
act  as  a  secondary  coil  of  a  transformer,  and  since  the  number  of 
turns  in  them  is  many  times  the  number  of  turns  on  the  armature, 
a  very  high  pressure  is  produced  in  them.  Secondly,  when  a  rotary 
is  started  up  in  this  manner  it  is  manifest  that  the  polarity  is  not 
definite,  and  this  is  a  very  important  point  in  view  of  parallel  work- 
ing. It  appears  then  that  this  method  of  starting  is  not  desirable. 
If  a  source  of  direct  current  is  available  we  may  use  this  to  drive 
the  rotary  up  to  the  speed  of  synchronism  as  a  direct  current  motor, 
the  parallelizing  being  conducted  as  described  in  the  chapter  on 
the  parallel  running  of  alternators.  In  many  cases  such  a  direct 
current  supply  is  available  from  a  storage  battery,  if  it  be  absent 
we  may  install  an  induction  motor  direct  coupled  to  a  small  direct 
current  dynamo  of  just  sufficient  power  to  run  the  rotaries  up  to 
speed. 

Pressure  regulation.  The  condition  that  is  required  to  be 
fulfilled  by  the  direct  current  circuit  of  a  rotary  is  in  general  that 
it  shall  supply  constant  pressure  at  its  terminals,  or  in  some  cases 
even  a  pressure  that  increases  with  the  load.  Since  the  value  of 
the  alternate  current  E.M.F.  is  definitely  related  to  the  direct  current 
one,  it  is  evident  that  the  necessary  increase  of  the  latter  as  the 
load  comes  on  could  be  provided  for  by  altering  the  alternate 
current  pressure  in  an  appropriate  manner.  This  is  in  some  cases 
done  by  supplying  the  different  phases  of  the  rotatry  with  suitable 
auto-transformers  provided  with  sets  of  terminals  giving  the  desired 
range  of  pressure  with  proper  intermediate  steps  in  the  manner 
described  in  Chap.  VI.  The  regulation  is  effected  by  means  of 
a  contact  arm  moving  over  the  terminals  of  the  successive  tappings, 
this  arm  being  commonly  actuated  by  a  small  induction  motor. 
Such  a  form  of  regulation  is  not  in  general  automatic. 

A  second  method  of  attaining  the  desired  effect  would  be 
obtained  by  winding  the  direct  current  circuit  as  a  compound 
wound  dynamo,  in  which  case  the  pressure  would  automatically 
rise  with  the  current  supplied  on  that  side.  Such  increase  of 
excitation  would  of  necessity  result  in  an  increase  of  the  back 
E.M.F.  of  the  machine  considered  as  a  synchronous  motor.  Consider 
the  case  shown  in  Fig.  248  where  any  of  the  lines  OG  represents 
the  constant  value  of  the  alternating  E.M.F.  on  any  phase  of  the 
motor.  Let  any  current  be  taken  as  given  by  the  line  MC,  and 
draw  as  before  the  triangle  MGQ  such  that  a  is  the  angle  whose 


THE    ROTARY   CONVERTER 


319 


tangent  is  the  ratio  of  the  reactance  of  the  armature  per  phase  to 
its  resistance.  Draw  the  circle  with  centre  G  and  radius  equal  to 
the  constant  applied  pressure.  When  the  current  and  this  pressure 
are  in  phase  the  pressure  line  will  be  given  by  OG  parallel  to  MC, 
the  corresponding  back  E.M.F.  of  the  rotary  being  OM.  With  the 


Fig.  248. 

same  current  it  will  be  readily  seen  that  a  small  increase  of  the 
rotary's  E.M.F.  consequent  on  an  increase  of  excitation  will  result 
in  the  triangle  OMG  becoming  O^MG,  and  in  this  case  the  current 
will  lead  the  terminal  pressure  as  shown  by  the  arrowheads  on  the 
current  and  pressure  vectors.  On  the  other  hand,  any  small 
diminution  in  the  E.M.F.  of  the  rotary  will  result  in  the  triangle 
becoming  as  shown  at  OMG,  and  the  current  will  lag  on  the 
pressure.  It  follows  that  if  the  current  and  pressure  are  in  phase 
for  any  definite  load,  an  angle  of  lead  or  lag  will  result  from  the 
alteration  of  excitation  consequent  on  any  alteration  of  the  con- 
ditions. The  loss  of  energy  in  heat  in  the  rotor's  armature 
depends,  as  we  have  seen,  on  the  load  and  this  phase  angle  between 
the  current  and  pressure,  and  for  any  definite  maximum  current 
will  be  less  the  phase  angle  whether  of  lead  or  lag.  It  follows 
that  it  is  most  suitable  to  arrange  so  that  the  phase  angle  is 
nearly  zero  when  the  load  is  a  maximum,  and  hence  with  less  loads 
the  current  will  in  this  case  lag  after  the  terminal  pressure. 

Consider  now  Fig.  249  in  which  the  angle  a  is  only  moderately 
large,  corresponding  to  a  rotary  with  comparatively  low  reaction. 
Let  MC  denote  the  full-load  current,  then  it  is  desired  that  the 


c, 


phase  angle  should  be  very  small,  or  that  the  corresponding 
terminal  pressure  should  be  given  by  OG  parallel  to  MC ;  hence 
the  rotary's  back  E.M.F.  will  be  given  by  OM.  Now  let  the  load 
be  very  small  as  shown  by  M&,  it  evidently  follows  from  the  con- 
struction that  the  corresponding  back  E.M.F.  is  given  by  OlMl,  and 


320  ALTERNATING   CURRENTS 

in  this  case  it  is  possible  that  the  latter  will  be  greater  than 
But  this  is  contrary  to  the  conditions  necessary  for  the  operation 
of  the  machine  as  a  compound  wound  direct  current  generator. 
Now  let  the  armature  have  a  much  higher  reactance  as  shown  in 


Fig.  250.  On  following  the  construction  indicated  by  the  same 
letters  as  the  last,  it  will  readily  be  seen  that  OM  can,  as  is 
required,  be  greater  than  MlOl.  Thus  if  the  rotary  has  a  proper 
amount  of  reaction  the  conditions  necessary  to  the  rise  of  E.M.F. 
with  load  with  a  desired  maximum  value  of  the  phase  angle 
between  current  and  terminal  pressure  can  be  fulfilled. 

Thus  the  compounding  can  be  successfully  carried  out  if  the 
proper  amount  of  reactance  is  provided  in  the  phases  of  the  rotary. 
The  reactance  of  the  machine  itself  may  be  supplemented  by  the 
provision  of  inductive  coils  placed  in  the  mains  leading  to  the 
slip-rings,  or  by  giving  the  transformers  supplying  those  rings  the 
proper  amount  of  leakage  field ;  in  the  chapter  on  transformers  it 
will  be  recollected  that  such  a  state  of  things  was  shown  to  be 
equivalent  to  the  presence  of  reactance  in  the  circuit. 

Efficiency.  For  small  machines  the  efficiency  can  be  measured 
by  the  direct  observation  of  the  input  and  output,  but  the  method 
is  of  little  utility  for  rotaries  of  any  size.  The  stray  power  methods 
must  then  be  used.  The  core  losses  and  other  losses  incident  to 
rotation  can  readily  be  found  by  running  the  machine  light  as 
a  direct  current  motor  at  its  normal  direct  current  pressure,  in 
such  a  case  the  power  being  observed  and  a  small  correction  made 
for  the  ohmic  loss  in  the  armature,  the  value  of  these  losses  is  at 
once  determined.  The  determination  of  the  armature  ohmic  loss 
can  be  calculated  from  the  measurement  of  the  armature  resistance 
and  the  use  of  the  expressions  found  for  the  relative  losses  of  the 
machine  as  a  direct  current  generator  and  as  a  rotary  with  different 
power  factors. 

A  combined  test  can  be  carried  out  on  the  lines  of  the  ordinary 
Kapp  test  of  two  similar  shunt  machines.  In  that  case  the 
machines  are  coupled  mechanically  together,  but  when  the 
machines  are  rotaries  this  coupling  can  be  effected  by  means  of 
the  alternate  current  sides,  as  shown  in  Fig.  251.  The  direct 
current  sides  are  connected  in  parallel  across  mains  that  are  put 
in  circuit  with  a  battery  or  other  source  of  direct  current  having 


THE   ROTARY   CONVERTER 


321 


the  proper  pressure  and  current  carrying  power,  and  the  current 
taken  and  the  pressure  are  read  by  the  ammeter  A  and  the 
voltmeter  V.  The  alternate  current  sides,  shown  as  three-phase, 
are  connected  by  means  of  three  auto-transformers  placed  in  series 
in  the  connecting  mains  and  with  the  other  ends  joined  in  star  as 


Fig.  251. 

shown.  Before  making  this  circuit  the  machines  are  brought  up 
to  synchronism  in  the  ordinary  way  with  lamps  (not  shown  in  the 
figure),  and  when  running  in  that  state  the  three-phase  sides  are 
connected.  By  alterations  in  the  tappings  on  the  three  auto- trans- 
formers and  by  suitable  regulation  of  the  fields  of  the  machines, 
any  desired  amount  and  character  of  load  can  be  circulated  between 
the  two.  The  amount  of  this  circulating  load  can  be  measured 
by  the  two  wattmeters  Wl  and  W2;  let  this  load  in  any  case  be 
denoted  by  W.  The  reading  of  the  ammeter  and  voltmeter  being 
C  and  E  the  total  power  that  is  being  supplied  is  evidently  EG. 
This  consists  of  the  actual  loss  in  the  machines  together  with  the 
loss  in  the  auto-transformers,  and  if  this  latter  has  been  previously 
determined  for  the  different  currents  taken,  the  nett  total  loss  in 
the  two  machines  can  be  deduced,  let  it  be  denoted  by  Wt.  Then, 
as  in  the  case  of  two  similar  dynamos,  if  this  loss  is  allotted  half  to 
each  machine,  and  if  the  efficiencies  are  taken  as  equal,  the  value 
of  the  efficiency  of  either  is  given  by 


-  £  "I 


Hunting.  Since  the  rotary  is  essentially  a  synchronous  motor 
the  phenomenon  of  hunting,  with  corresponding  surging  in  the 
current  received  from  the  alternating  mains,  will  be  liable  to  occur, 
but  since  the  output  is  electrical  and  not  mechanical  energy,  such 
surging  will  produce  effects  concerned  with  the  collection  of  the 
direct  current.  Thus  wrhen  the  machine  is  slowing,  it  will  be 
receiving  more  than  the  normal  current,  and  while  accelerating  it 


L. 


21 


322  ALLERNATING  CURRENTS 

will  receive  a  different  one ;  similar  variations  will  occur  on  the 
direct  current  side.  Hence  any  distorting  effect  due  to  the  direct 
current  on  the  poles  will  thus  vary  in  amount,  and  hence  the 
•commutating  field  at  the  brushes  will  also  vary.  It  may  result 
that  conditions  of  sparkless  commutation  are  no  longer  fulfilled 
during  some  portions  of  the  hunting  period.  It  is  thus  of  great 
importance  that  such  surges  should  be  rapidly  damped  out,  and 
consequently  amortisseurs  should  be  fitted.  It  is  especially 
important  to  damp  out  any  variations  in  the  commutation  fields, 
a,nd  the  copper  pole  rings  are  often  made  especially  heavy  for  that 
purpose :  in  some  cases  extra  short-circuited  grids  are  put  between 
the  polar  horns  as  well  as  on  the  poles. 

In  the  case  of  a  monophase  rotary  it  is  evident  that  the  power 
received  being  no  longer  constant  as  on  the  polyphase  machine, 
but  becoming  at  least  zero  twice  per  alternation,  and  in  general 
becoming  negative,  the  rotary  must  act  as  a  generator  for  those 
intervals  and  hence  there  is  bound  to  be  a  rapid  to  and  fro 
oscillation  of  the  field  of  the  machine  in  the  air  gap.  This  can 
be  greatly  diminished  by  the  amortisseur,  but  it  will  be  seen 
that  satisfactory  working  is  much  more  difficult  to  attain  in  this 
case. 

Motor  generator.  The  production  of  direct  currents  from 
alternating  may  of  course  be  carried  out  by  an  ordinary  motor 
generator  consisting  of  an  induction  motor  coupled  to  a  direct 
current  machine.  Such  an  arrangement  has  the  disadvantage  of 
being  less  efficient  than  the  rotary  converter  owing  to  the  extra 
transformation  of  energy  that  is  in  general  involved,  also  in  such  a 
case  the  power  factor  is  necessarily  less  than  unity  and  in  general 
less  than  can  readily  be  attained  with  a  rotary.  The  arrangement 
has,  however,  certain  advantages  in  regard  to  the  greater  flexibility 
of  the  pressure  ratios  between  the  two  sides  and  is  in  general  some- 
what easier  to  deal  with  in  the  manner  of  starting,  etc.  The  choice 
of  method  must  be  made  with  regard  to  the  special  circumstances 
of  each  case. 


INDEX. 


Air  core  transformer,  80 
Air  gap,  effect  on  choking  coil,  51 
Alternating  E.M.F.,  production  of,  1 
Alternator,  combined  test,  138 

on  constant  load,  248 

deceleration  test,  139 

efficiency,  136 

induction  curve,  103 

loss  of  energy  in,  135 

mean  E.M.F.,  103 

no-load  test,  137 

nominal  induced  E.M.F.,  243 

short  circuit  test,  137 
Amortisseur,  305 
Armature,  back  reactance,  263 

cross  reactance,  263 

reactance,  polyphaser,  260 

reaction,  detail,  258 

true  reactance  of,  242 
Asynchronous  generator,  190 
Auto-transformer,  78 

B.-H.  curve  for  iron,  47 

Back  pressure,  16 

Balance,  146 

Beats,  279 

Breadth  coefficient,  105 

Cascade  working  of    induction  motors, 
228 

Characteristic,  external,  244 

locked,  of  induction  motor,  206 
mechanical,  of  induction  motor,  187 
mechanical,  of   monophase   motor, 

225  . 

open  circuit,  of  alternator,  243 
short  circuit,  of  alternator,  243 

Choking  coil,  46 

Combined  test,  alternator,  138 
rotary,  320 
transformer,  87 

Common     Iron     Core     in     Polyphase 
Transformer,  162 

Commutator  for  small  phase  angles,  221 

Compensator,  78 

Compounding,  pole  piece,  256 
transformer,  251 

Concatenation,  229 


Core  loss,  48 

Cranes,  induction  motor  for,  188 
Current-transformer,  80 
Cycle  of  core,  determination  of,  experi- 
mentally, 131,  133 

Degrees,  electrical,  111 
Distributed  winding,  108 
Dynamometer,  electric,  5 

Eddy  currents,  loss  due  to,  53 
Effective  pressure,  16 
Efficiency  of  alternator,  136 

induction  motor,  222 

rotary  converter,  320 

transformer,  85,  86,  87 
Electrometer,  use  as  wattmeter,  40 
Electrostatic  voltmeter,  8 
E.M.F.,  nominal  induced,  of  alternator, 

243 
Equivalent  simple  harmonic  current  and 

pressure,  29 
External  characteristic,  244 

Form  Factor,  120 

Harmonic  quantities,  resolution  of,   12 

summation  of,  12 

representation  of,  2 

simple,  E.  M.F.,  2 
Harmonics,  114 

absence  of  even,  116 

effect  on  current,  117 

power  due  to,  120 
Hemitropic  winding,  111 
Heyland  diagram,  191,  195 

experimental  determination  of,  198 

slip,  193,  197 

torque,  193,  197 
Hot  wire  instruments,  7 
Hunting,  277 

rotary  converter,  321 
Hysteresis     test    of     iron,     wattmeter 

method,  91 

Hysteretic  lead,  angle  of,  48,  51 
Hysteretic  loss  and  induction,  curve  of» 
50 


324 


INDEX 


Impedance,  definition,  17 

equivalent,  of  induction  motor,  207 

equivalent,  of  transformer,  68 

synchronous,  244 

triangle,  19 

Impressed  pressure,  definition,  16 
Induction  curve  of  alternator,  103 
Induction  motor 

for  cranes,  188 

concatenation  or  cascade,  228 

dynamo  action,  188 

effect  of  want  of  phase  balance,  205 

efficiency,  222 

equivalent  reactance,  207 

equivalent  resistance,  207 

fractional  slip,  207 

ideal  winding,  173 

the  induced  E.M.F.,  175 

leakage  fluxes,  178 

locked  characteristic,  206 

mechanical  characteristic,  187 

multipolar  fields,  218 

no-load  test,  198,  203 

output,  expression  for,  210 

phase  relations  of  currents,  176 

practical  form  of,  190 

prediction  of  performance  of,  213 

representation  of  flux  belt,  176 

running  condition,  184 

slip,  168 

standstill  test,  198,  206 

starting  condition,   185 

starting  of,  219 

stator  current,  expression  for,  211 

synchronous  speed  of,  169 

torque,  expression  for,  183,  209 

vector  diagram,  180 
Instantaneous     curves,     determination 

of,  123 

Integral  of  vector  quantity,  representa- 
tion of,  13 

Lag,  angle  of,  10 
Lead,  angle  of,  10 
Leakage  Flux,  armature,  242 

induction  motor,  178 

transformer,  61 

Mean  value  of  an  alternating  quantity, 

4 

Mean  values,  theorem  on,  33 
Mesh  connection,  150 
Meters,  rotating  field,  230 

polyphase,  238 

quadrature  devices,  235-6 

sliding  field,  233 

torque,  231 
Monophase  motor,  224 

fluxband  in,  225 

mechanical  characteristic,  225 

starting,  227 
Motor,  induction,  see  Induction  motor, 

monophase,  224 


Motor,  series,  94 

synchronous,  282 
Motor-generator,  322 
Multipolar  fields  in  alternator,  109 

in  induction  motor,  218 

Neutral  point,  147,  313 
Neutral  point  resistance,  152 
No-load  test  alternator,  137 

induction  motor,  198,  203 

transformer,  72 
Nominal  induced  E.M.F.,  243 

Open  circuit  characteristic  of  alterna- 
tor, 243 

Open  circuit  test,  see  No-load  test 
Oscillations,  see  Synchronous  motor,  299 
Oscillograph,  127 

Parallel    working    of    alternators,   con- 
ditions, 273 

Parallelizing,  process  of,  277 
Periodic  time,  1 
Periodicity  measurer,  9 
Phase  angle,  maximum  value  in  choking 
coil,  56 

measurement  of  small,  39,  221 
Phase  meter,  238 
Phase  transformations,  158 
Pitch  of  poles,  110 
Point  to  point  method,  123 
Power,  apparent,  29 

fictitious,  45 

mean,  28 

electrometer,  measure  of,  40 

three     voltmeter     and      ammeter 
methods,  38 

polyphase  circuit,  152 

see  also  Wattmeter 
Power  component  of  current,  28 
Power  factor,  29 

condition  of  unit,  31 

Kate  of  change  of  vector,  representation, 

13 
Reactance  (definition),  17 

back,  of  armature,  263 

cross,  of  armature,  263 

equivalent,  in  induction  motor,  207 

pressure,  16 

synchronous,  242 

true,  of  armature,  242 
Eeaction  of  armature,  240 
Kegulation  of  transformer,  59 
Eesistance,     equivalent,     in     induction 
motor,  207 

equivalent,  in  transformer,  67 
Eesonance,  18,  119 
Rotary  converter,  310 

current  relations,  314 

hunting,  321 

efficiency,  320 

E.M.F.  relations,  310 


INDEX 


325 


Rotary  converter 

ohmic  loss.  315 

pressure  regulation,  318 

starting,  317 
Rotating  field,  164 

Rotating    field    motor,    see    Induction 
motor,  167 

Saturation  curve,  243 
Series  A.  c.  motor,  94 
Short   circuit    characteristic    of    alter- 
nator, 243 
Short  circuit  test  of  alternator,  137 

transformer,  68 

Slip,  see  Induction  motor,  168,  207 
Star  connection,  149 
Starting,  induction  motor,  219 

monophase  motor,  227 

rotary  converter,  317 
Stray  power  method,  86 
Surging,  277 

Synchronous  impedance,  244 
Synchronous  motor,  247,  282 

maximum  output,  289 

motor  E.M.F.,  construction,  292 

motor  E.M.F.  for  maximum  power 
supplied,  290 

oscillations,  free,  damped,  303 

oscillations,  free,  undamped,  299 

oscillations,  forced,  307 

power,  285 

"V"  curves,  294,  298 
Synchronous  reactance,  242 

Temperature,  effect  on  iron,  90 
Three-phase  dynamo,  147 
Time,  positive  value  of,  3 
Torque  with  rotating  field,  167 

induction  motor,  183,  193,  197,  209 
Transformer,  56 


Transformer,  air  core,  80 

combined  test,  87 

common  core,  162 

for  current,  80 

efficiency,  85 

equivalent  impedance,  68 

equivalent  resistance,  67 

leakage  flux,  61 

leakage  equivalent  to  reactance,  67 

losses,  85 

no-load  test,  72 

open  circuit  test,  71 

polyphase,  156 

primary  of,  56 

ratio  of  transformation,  59 

regulation,  59 

regulation  and  leakage,  65 

rejection  of  waste,  90 

secondary  of,  56 

short  circuit  test,  68 

six-phase,  160 

stray  power  method,  86 

twelve-phase,  161 

two-  to  three-phase,  158 
Two-phase  dynamo,  143 

Unbalanced  load,  measure  of,  154 

"V"  curves,  294,  298 

Virtual  value  of  an  alternating  quantity, 

definition,  4 
complex  E.M.F.,  119 

Wattless  component  of  current,  28 
Wattmeter  (ordinary  type),  34 

compensator  in,  35 

determination  of  constant,  35 

error,  35 

iron  cored,  81 

rotating  field,  238 


CAMBRIDGE  :    PRINTED    BY   JOHN    CLAY,    M.A.    AT   THE    UNIVERSITY   PRESS. 


066 


203809 


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